Author name: Prasanna

These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-11-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise

Question 1.
Show that the line joining the origin to the point (2,1,1) is perpendicular to the line de-termined by the points (3, 5, – 1), (4, 3, -1).
Solution:
The direction ratios of the line joining (0, 0, 0) and (2, 1, 1) are 2, 1,1 The direction ratios Of the line joining (3, 5, -1) and (4, 3, -1) are 1, -2, 0
∴ a₁a₂ + b₁b₁ + C₁C₂
= 2(1) + 1(-2) + 1 (0) = 0
The lines are perpendicular.

Question 2.
If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂ – m₂ n₁, n₁ l₂ – n₂ l₁, l₁ m₂ – l₂ m₁
Solution:.
Let \(\hat{a}\) and \(\hat{b}\) be unit vectors in the direction of the lines
∴ \(\hat{a}=l_{1} \hat{i}+m_{1} \hat{j}+n_{1} \hat{k}\)
\(\hat{b}=l_{2} \hat{i}+m_{2} \hat{j}+n_{2} \hat{k}\)
Since a and b are mutually perpendicular, then \(\hat{a}\) x \(\hat{b}\) is a unit vector perpendicular to \(\hat{a}\) and \(\hat{b}\)

∴ The direction cosines are m₁n₂ – m₂ n₁, n₁ l₂ – n₂ l₁, l₁ m₂ – l₂ m₁

Question 3.
Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.
Solution:
a₁ = a, b₁ = b, c₁ = c
a₂ = b – c, b₂ = c – a, c₂ = a – b
Let θ be the angle between the lines

Question 4.
Find the equation of a line parallel to x- axis and passing the origin.
Solution:
The direction ratios of any line parallel to x-axis is 1, 0, 0. Hence the equation of the line passing through the origin having direction ratios 1,0,0 is
\(\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\)
i.e., \(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\)

Question 5.
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (- 4, 3, -6) and (2, 9, 2) respectively, then find the angle between the line AB and CD.
Solution:
The direction ratios of AB
= 4 – 1, 5 – 2, 7 – 3 = 3, 3, 4
The direction ratios of CD = 6, 6, 8
Let θ be the angle between AB and CD

Question 6.
If the lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\) are perpendicular, find the value of k.
Solution:
The equation of the lines are
\(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and
\(\frac{x-1}{3 k}=\frac{y-1}{1}=\frac{z-6}{-5}\)
The direction ratios of 1^st line = – 3, 2k, 2
The direction ratios of II^nd line = 3k, 1, – 5
Since the two lines are perpendicular,
a₁a₂ + b₁b₂ + c₁c₂ = 0
– 3(3k) + 2k(1) + 2(- 5) = 0
– 9k + 2k – 10 = 0
– 7k = 10
k = \(\frac { -10 }{ 7 }\)

Question 7.
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \(\vec{r} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})+9\) = 0.
Solution:
Let \(\hat{a}\) be the position vector of the point (1, 2, 3)
a = \(\hat{i}+2 \hat{j}+3 \hat{k}\)
The direction ratios of the normal to the plane are 1, 2, – 5.
Let b = \(\hat{i}+2 \hat{j}-5 \hat{k}\)
Let \(\hat{r}\) be the position vector of any point on the line.
The vector equation is \(\vec{r}=\vec{a}+\lambda \vec{b}\)
i.e., \(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+2 \hat{j}-5 \hat{k})\)

Question 8.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 2
Solution:
The given plane is \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 2
i.e., x + y + z – 2
The equation of any plane parallel to the given plane is x +y + z = k … (1)
Since the required plane passes through the point (a, b, c), we get a + b + c = k
Thus (1) gives the required plane as x + y + z = a + b + c

Question 9.
Find the shortest distance between the lines \(\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})\) and \(\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})\)
Solution:

Question 10.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
Solution:
The equation of the line through the points (5,1, 6) and (3,4, 1) is
\(\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}=\lambda\)
i.e., \(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda\)
i.e., x = – 2λ. + 5, y = 3λ + 1, z = – 5λ + 6
Since this line crosses the YZ – plane, x coordinate is zero
i.e., – 2λ + 5 = 0 ∴ λ = \(\frac { 5 }{ 2 }\)
\(y=3\left(\frac{5}{2}\right)+1=\frac{17}{2}\)
\(z=-5\left(\frac{5}{2}\right)+6=\frac{-13}{2}\)
∴ The required point is \(\left(0, \frac{17}{2}, \frac{-13}{2}\right)\)

Question 11.
Find the coordinates of the point where the line through (5, 1,6) and (3,4, 1) crosses the ZX- plane.
Solution:
The equation of the line through (5, 1, 6) and (3, 4, 1) is
\(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda\)
i.e., x = – 2λ + 5, y = 3λ + 1, z = – 5λ + 6
Since this line crosses the ZX – plane,
y coordinate is zero.

Question 12.
Find the coordinates of the point where the line through (3, – 4, -5) and (2, -3, 1) crosses the plane 2x + y + z = 7.
Solution:
The equation of the line through (3, – 4, – 5) and (2, – 3, 1) is

The coordinate of any point P on the line is (- λ + 3, λ – 4, 6λ – 5)
P is a point on the plane 2x + y + z = 7
i.e., 2(- λ + 3) + (λ – 4) + (6λ – 5) = 7
– 2λ + 6 + λ – 4 + 6λ – 5 = 7
5λ – 3 = 7 ∴ λ = 2
∴ P is (- 2+ 3, 2 – 4, 6(2) – 5) = (1, – 2, 7)
∴ The coordinates of the point which crosses the plane is (1, -2, 7)

Question 13.
Find the equation of the plane passing through the point (- 1, 3, 2) and perpendicular to each of the planes x + 2y +3z = 5 and 3x + 3y + z = 0.
Solution:
The plane passes through (- 1, 3, 2)
∴ x₁ = – 1, y₁ = 3, z₁ = 2
The direction ratios of the normal to the planes are 1, 2, 3 and 3, 3, 1.
Hence the equation of the plane is
\(\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|\) = 0
\(\left|\begin{array}{ccc} x+1 & y-3 & z-2 \\ 1 & 2 & 3 \\ 3 & 3 & 1 \end{array}\right|\) = 0
– 7(x + 1) + 8(y – 3) – 3(z – 2) = 0
i.e., – 7x – 7 + 8y – 24 – 3z + 6 = 0
– 7x + 8y – 3z – 25 = 0
7x – 8y + 3z + 250 = 0

Another method:
The equation of a plane passing through the point(- 1, 3, 2) is
a(x + 1) + b(y – 3) + c(z – 2) = 0 … (1)
where a, b and c are the direction ratios of the normal to the plane.
Since (1) is perpendicular to the planes
x + 2y + 3z = 5 and 3x + 3y + z = 0, we get
a + 2b + 3c = 0 … (2) and
3a + 3b + c = 0 … (3)
∴ By the rule of cross multiplication, we get
\(\frac{a}{\left|\begin{array}{ll} 2 & 3 \\ 3 & 1 \end{array}\right|}=\frac{0}{\left|\begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array}\right|}=\frac{c}{\left|\begin{array}{ll} 1 & 2 \\ 3 & 3 \end{array}\right|}\)
i.e., \(\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}\)
Since a, b, c are the direction ratios of the normal to the plane, we can take a = – 7, b = 8, c = – 3.
∴ (1) → 7(x+ 1) + 8(y – 3) – 3(z – 2) = 0
i.e., – 7x + 8y – 3z – 25 = 0
or 7x – 8y + 3z + 25 = 0

Question 14.
If the points (1, 1, p) and (-3, 0, 1) are equidistant from the plane \(\vec{r} \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})+13\) = 0, then find the value of p.
Solution:
The equation of the plane is,

Question 15.
Find the equation of the plane passing through the line of intersection of the planes \(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 1 and \(\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4\) = 0 and parallel to x- axis.
Solution:
Equation of the given planes are
x + y + z – 1 = 0 and 2x + 3y – z + 4 = 0
The equation of the plane passing through the line of intersection of the planes is
(x + y + z – 1) + (2x + 3y – z + 4) = 0
(1 + 2λ)x + (1 + 3λ)y + (1 – λ )z + (4 – 1) = 0 … (1)
The direction ratios of the normal to the plane are 1 + 2λ , 1 + 3λ , 1 – λ
The direction ratios of x-axis are 1, 0, 0
Since the required plane is parallel to the x-axis, the normal to the plane is perpendicular to x-axis.

Question 16.
If O is the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.
Solution:
\(\overrightarrow{\mathrm{OP}}=\hat{i}+2 \hat{j}-3 \hat{k}\)
Direction ratios of OP are 1, 2, -3
OP is perpendicular to the plane
∴ Direction ratios of the normal to the plane
are a = 1, b = 2, c = -3
P( 1, 2, – 3) is a point on the plane.
Equation of the plane is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
i.e, 1(x – 1) + 2(y – 2) – 3(z + 3) = 0
x – 1 + 2y – 4 – 3z – 9 = 0
x + 2y – 3z – 14 = 0
The equation of the plane is x + 2y – 3z – 14 = 0

Question 17.
Find the equation of the plane which contains the line of intersection of the planes \(\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4\) = 0, \(\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5\) = 0 and which is perpendicular to the plane \(\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8\) = 0.
Solution:
The equation of the given planes are x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
The equation of the plane containing the line of inÍersection of the above plane is
(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0
(1 + 2λ)x + (2 + λ)y (3 – λ)z + (5λ – 4) = 0 … (1)
Since (1) is perpendicular to the plane
5x + 3y – 6z + 8 = 0, we get
5(1 + 2λ) + 3(2 + λ) – 6(3 – λ) = 0
5 + 10λ + 6 + 3λ – 18 + 6λ = 0
19λ – 7 = 0
∴ λ = \(\frac { 7 }{ 19 }\)

Question 18.
Find the distance of the point (- 1, – 5, – 10) from the point of intersection of the line \(\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})\) = 5
Solution:
The cartesian equation of the line \(\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})\) is
\(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}\) = λ … (1)
The cartesian equation of the plane
\(\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})\) is x – y + z = 5 … (2)
The coordinate of any point on this line is P(2 + 3λ, – 1 + 4λ, 2 + 2λ).
Let the line intersects the plane at P, then P is a point on the plane.
∴ (2) → (2 + 3λ) – (-1 + 4λ) + (2 + 2λ) = 5
2 + 3λ + 1 – 4λ + 2 + 2λ = 5
∴ λ = 0
∴ P is (2, – 1,2) and the given point is Q(- 1, – 5, – 10)
∴ PQ = \(\sqrt{(-1-2)^{2}+(-5–1)^{2}+(-10-2)^{2}}\)
= \(\sqrt{(-3)^{2}+(-4)^{2}+(-12)^{2}}\)
= \(\sqrt{9+16+144}=\sqrt{169}\)
= 13

Question 19.
Find the vector equation of the line passing through (1,2,3) and parallel to the planes \(\vec{r} \cdot(\hat{i}-\hat{j}+2 \hat{k})=5 \text { and } \vec{r} \cdot(3 \hat{i}+\hat{j}+\hat{k})=6\)
Solution:
The planes are \(\hat{r}\). \(\hat{n}\)₂ = d₁
and \(\hat{r}\). \(\hat{n}\)₂ = d₂
where \(\hat{n}\)₁ = \(\hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{n}_{2}=3 \hat{i}+\hat{j}+\hat{k}\)
Since the required line is parallel to the planes, it is perpendicular to both \(\hat{n}\)₁ and \(\hat{n}\)₂.
\(\hat{n}\)₁ x \(\hat{n}\)₂ is perpendicular to both \(\hat{n}\)₁ and \(\hat{n}\)₂
\(\vec{n}_{1} \times \vec{n}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & 2 \\
3 & 1 & 1
\end{array}\right|=-3 \hat{i}+5 \hat{j}+4 \hat{k}\)
The required line passes through the point (1, 2, 3) and is parallel to \(\hat{n}\)₁ x \(\hat{n}\)₂.
∴ The equation of the line is
\(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-3 \hat{i}+5 \hat{j}+4 \hat{k})\)

Question 20.
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:
\(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\)
\(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\)
Solution:
Let \(\vec{b}_{1}\) and \(\vec{b}_{2}\) be the direction of the lines where \(\vec{b}_{1}=3 \hat{i}-16 \hat{j}+7 \hat{k}\) and \(\vec{b}_{2}=3 \hat{i}+8 \hat{j}-5 \hat{k}\).
The required line is perpendicular to both \(\vec{b}_{1}\) and \(\vec{b}_{2}\). Hence it is parallel to \(\vec{b}_{1}\) x \(\vec{b}_{2}\)
\(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -16 & 7 \\
3 & 8 & -5
\end{array}\right|=24 \hat{i}+36 \hat{j}+72 \hat{k}\)
The direction ratios of \(\vec{b}_{1}\) x \(\vec{b}_{2}\) = 24, 36, 72
= 2, 3, 6
The required line passes through the point (1, 2, – 4 ) and has direction ratios 2, 3, 6.
Hence the equation is
\(\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)

Question 21.
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then \(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}\).
Solution:
The equation of a plane in the intercept from is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1.

Question 22.
Distance between the two planes:
2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
a. 2 units
b. 4 units
c. 8 units
d. \(\frac{2}{\sqrt{29}}\) units
Solution:
d. \(\frac{2}{\sqrt{29}}\) units

Hence the planes are parallel.
The distance between the parallel planes is the difference between their distance from the origin.
The distance of the plane 2x + 3y + 4z = 4

Question 23.
The planes 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
a. perpendicular
b. parallel
c. intersect y-axis
d. passes through (0, 0, \(\frac { 5 }{ 4 }\))
Solution:
b. parallel

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 10 Vector Miscellaneous Exercise

Question 1.
Write down a unit vector in XY – plane, making an angle of 30° with the positive direction of x – axis.
Solution:
Let P(x, y) be any point on the line making 30° with x-axis. Let \(\vec{a}\) = x\(\hat{i}\) + y\(\hat{i}\) be a unit vector.

Another method:
The line in the above figure lies in X Y plane and makes 30° with positive direction of x – axis. Hence the line makes 60° with y- axis and 90° with z – axis.
The direction cosines of the line
= cos 30°, cos 60°, cos 90° = \(\frac{\sqrt{3}}{2}\), \(\frac { 1 }{ 2 }\), 0
Hence the unit vector is \(\frac{\sqrt{3}}{2}\)\(\hat{i}\) + \(\frac { 1 }{ 2 }\)\(\hat{j}\) + 0\(\hat{k}\).

Question 2.
Find the scalar components and magnitude of the vector joining the points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂).
Solution:

Question 3.
A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.
Solution:

Let the girl starts from O and stops at P. Let the coordinates of P be (x, y).
\(\vec{r}\) = \(\overrightarrow{OP}\)
∴ \(\vec{r}\) = – x\(\hat{i}\) + y \(\hat{j}\) … (1)
In ∆ QMP, sin60° = \(\frac { y }{ 3 }\)

Question 4.
If \(\vec{a}\) = \(\vec{b}\) + \(\vec{c}\), then is it true that |\(\vec{a}\)| = |\(\vec{b}\)| + |\(\vec{c}\)|? Justify your answer.
Solution:

Question 5.
Find the value of x for which x\((\hat{i}+\hat{j}+\hat{k})\) is a unit vector.
Solution:

Question 6.
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}+\hat{k}\).
Solution:
\(\vec{a}+\vec{b}\) be the resultant of \(\vec{a}\) and \(\vec{b}\).

Question 7.
If \(\vec{a}\) = \(\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}\) = \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{c}\) = \(\hat{i}-2 \hat{j}+\hat{k}\), find a unit vector parallel to the vector \(2 \vec{a}-\vec{b}+3 \vec{c}\).
Solution:

Question 8.
Show that the points A (1, – 2, – 8), B (5, 0, – 2)and C (11, 3, 7) are collinear and find the ratio in which B divides AC.
Solution:

\(\overrightarrow{AB}\) is a scalar multiple of \(\overrightarrow{BC}\)
∴ \(\overrightarrow{AB}\) || \(\overrightarrow{BC}\) and B is the common point.
Hence A, B, C are collinear.
Also \(\frac{|\overrightarrow{\mathrm{AB}}|}{|\overrightarrow{\mathrm{BC}}|}=\frac{2}{3}\)
∴ The point B divides AC in the ratio 2 : 3.

Another method:
Let \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) be the position vectors of A, B and C.
∴ \(\vec{a}\) = \(\hat{i}-2 \hat{j}-8 \hat{k}\), \(\vec{b}\) = \(5 \hat{i}+0 \hat{j}-2 \hat{k}\), \(\vec{c}\) = \(11 \hat{i}+3 \hat{j}+7 \hat{k}\)
Let B divide AC in the ratio λ : 1
By section formula, the position vector of

Equating components of \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\), we get

∴ λ = \(\frac{2}{3}\) satisfies (1) and (3)
Hence A, B and C are collinear and B divides AC in the ratio \(\frac{2}{3}\) : 1 That is, in the ratio 2 : 3.

Question 9.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2\(\vec{a}\) + \(\vec{b}\)) and (\(\vec{a}\)-3\(\vec{b}\)) externally in the ratio 1 : 2. Also, show that P is the midpoint of the line, segment RQ.
Solution:
\(\overrightarrow{\mathrm{OP}}=2 \vec{a}+\vec{b}, \overrightarrow{\mathrm{OQ}}=\vec{a}-3 \vec{b}\),
R divides PQ in the ratio 1 : 2 externally.

∴ P is the midpoint of the line segment RQ.

Question 10.
The two adjacent sides of a parallelogram are \(2 \hat{i}-4 \hat{j}+5 \hat{k} \text { and } \hat{i}-2 \hat{j}-3 \hat{k}\). Find the unit vector parallel to its diagonal. Also, find its area.
Solution:
Let \(\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}\) be any two adjacent sides of a parallelogram.
\(\vec{a}\) + \(\vec{b}\) is a vector along the diagonal

Question 11.
Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\).
Solution:
Let a, P and y be the angles made by the vector with the axes OX, OY and OZ.
Let l, m and n be the direction cosines.
Since the vector is equally inclined to the
axes, α = ß = γ
∴ l = cosα, m = cos α, n = cos α
We have l² + m² + n² = 1
i.e, cos² α + cos² α + cos² α = 1
⇒ 3 cos² α = 1
cos² α = \(\frac { 1 }{ 3 }\)
cos α = \(\frac{1}{\sqrt{3}}\)
∴ l = m = n = \(\frac{1}{\sqrt{3}}\)
Hence the direction cosines of the vector equally inclined to the axes are \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\).

Question 12.
Let \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\) and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\) Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{c}\). \(\vec{d}\) = 15.
Solution:
Let \(\vec{d}\) = \(x \hat{i}+y \hat{j}+z \hat{k}\)
Since \(\vec{d}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\)
\(\vec{d}\).\(\vec{a}\) = 0 and \(\vec{d}\).\(\vec{b}\) = 0.
x + 4y + 2z = 0 … (1) and
3x – 2y + 7z = 0 … (2)
Given \(\vec{c}\).\(\vec{d}\) = 15 … (3)
2x – y + 4z = 15 … (3)
Solving (1), (2) and (3), we get x = \(\frac { 160 }{ 3 }\)
y = \(\frac { – 5 }{ 3 }\) and z = \(\frac { -70 }{ 3 }\)
∴ \(\vec{d}=\frac{160}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{70}{3} \hat{k}\)
\(\vec{d}=\frac{1}{3}(160 \hat{i}-5 \hat{j}-70 \hat{k})\)

Question 13.
The scalar product of the vector \(\hat{i}+\hat{j}+\hat{k}\) with a unit vector along the sum of vectors \(2 \hat{i}+4 \hat{j}-5 \hat{k} \text { and } \lambda \hat{i}+2 \hat{j}+3 \hat{k}\) is equal to one. Find the value of λ.
Solution:

Question 14.
If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular vectors of equal magnitudes, show that the vectors \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).
Solution:
Given that \(|\vec{a}|=|\vec{b}|=|\vec{c}|=\lambda\) … (1)
Since \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular

Hence \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined to \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\).

Question 15.
Prove that \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}\), if and only if \(\vec{a}\), \(\vec{b}\) are perpendicular, given a ≠ \(\vec{0}\), b ≠ \(\vec{0}\)
Solution:
\((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
\((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}+2(\vec{a} \cdot \vec{b}) \ldots \ldots\) … (1)
Case 1
Let
From (1), we get \(\vec{a}\), \(\vec{b}\) = 0
Hence \(\vec{a}\) ⊥ \(\vec{b}\) since \(\vec{a}\) and \(\vec{b}\) are nonzero vectors.

Case 2
Let \(\vec{a}\) ⊥ \(\vec{b}\). Then \(\vec{a}\).\(\vec{b}\) = 0
(1) → \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}\) + 0
\((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}\)

Question 16.
If θ is the angle between two vectors \(\vec{a}\) and \(\vec{b}\), then \(\vec{a}\).\(\vec{b}\) ≥ 0 only when
a. 0 < θ < \(\frac { π }{ 2 }\)
b. 0 ≤ θ ≤ \(\frac { π }{ 2 }\)
c. 0 < θ < π
d. 0 ≤ θ ≤ π
Solution:

Question 17.
Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors and θ is the angle between them. Then \(\vec{a}\) + \(\vec{b}\) is a unit vector if
a. θ = \(\frac { π }{ 4 }\)
b. θ = \(\frac { π }{ 3 }\)
c. θ = \(\frac { π }{ 2 }\)
d. θ = \(\frac { 2π }{ 3 }\)
Solution:

Question 18.
The value of
\(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j}) \text { is }\)
a. 0
b. -1
c. 1
d. 3
Solution:
c. 1
\(\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j}) \text { is }\)
= \(\hat{i} \cdot \hat{i}-\hat{j} \cdot \hat{j}+\hat{k} \hat{k}\)
= 1 – 1 + 1 = 1

Question 19.
If θ is the angle between any two vectors \(\vec{a}\) and \(\vec{b}\), then \(|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|\), when θ is equal to
a. 0
b. \(\frac { π }{ 4 }\)
c. \(\frac { π }{ 2 }\)
d. π
Solution:
b. \(\frac { π }{ 4 }\)

These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2/

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2

Ex 11.2 Class 12 NCERT Solutions Question 1.
Show that the three lines with direction cosines:
\(\frac { 12 }{ 13 } ,\frac { -3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 4 }{ 13 } ,\frac { 12 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 12 }{ 13 } \) are mutually perpendicular.
Solution:
Consider the lines with direction cosines
\(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} \text { and } \frac{4}{13}, \frac{12}{13}, \frac{3}{13}\)
The lines with direction cosines l₁, m₁, n₁ and l₂, m₂, n₂ are perpendicular if
l₁l₂ + m₁m₂ + n₁n₂ = 0
l₁l₂ + m₁m₂ + n₁n₂
= \(\left(\frac{12}{13} \times \frac{4}{13}\right)+\left(\frac{-3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{3}{13}\right)\)
= \(\frac{48}{169}-\frac{36}{169}-\frac{12}{169}=0\)
∴ The lines are perpendicular.
Consider the lines with direction cosines

are mutually perpendicular.

Class 12 Math Ex 11.2 NCERT Solutions Question 2.
Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Solution:
Let A (1, – 1, 2), B(3, 4, – 2), C (0, 3, 2) and D(3, 5, 6) be the points
Direction ratios of AB = 3 – 1, 4 – 1, – 2 – 2 = 2, 5, – 4
Direction ratios of CD = 3 – 0, 5 – 3, 6 – 2 = 3, 2, 4
a₁a₂ + b₁b₂ + c₁c₂ = 2(3) + 5(2) + (- 4) (4) = 6 + 10 – 16 = 0
Hence AB ⊥ CD.

Ex 11.2 12 Class NCERT Solutions Question 3.
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (- 1, – 2, 1) and (1, 2, 5).
Solution:
Let A(4, 7, 8), B (2, 3, 4), C (- 1, – 2, 1) and D(1, 2, 5) be the points
Direction ratios of AB = 2 – 4, 3 – 7, 4 – 8 = – 2, – 4, – 4
Direction ratios of CD = 1 – 1, 2 – 2, 5 – 1 = 2, 4, 4

Question 4.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \(3\hat { i } +2\hat { j } -2\hat { k }\)
Solution:
Let \(\vec{a}\) be the position vector of the point (1, 2, 3) and \(\vec{b}\) be the vector \(3 \hat{i}+2 \hat{j}-2 \hat{k}\)
∴ \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)
The vector equation of the line is \(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)
\(\vec{r}\) = (∴ \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)) + λ(\(3\hat { i } +2\hat { j } -2\hat { k }\))

Question 5.
Find the equation of the line in vector and in cartesian form that passes through the point with position vector \(2\hat { i } -\hat { j } +4\hat { k }\) and is in the direction \(\hat { i } +2\hat { j } -\hat { k }\).
Solution:
Let \(\vec{a}\) be the position vector of the point \(\vec{a}\) = \(\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}\)
\(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\)
The vector equation of a line passing through the point with position vector \(\vec{a}\) and parallel to \(\vec{b}\) is \(\vec{r}\) = \(\vec{a}+\lambda \vec{b}\)
Hence the equation of the line is
\(\vec{r}=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})\)
The cartesian equation is
\(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}\)

Another Method:
The point is (2, – 1, 4)
The direction ratios of the line are 1, 2, – 1
The equation of the line passing through (x₁, y₁, z₁) and having direction ratios a, b,c is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
Hence cartesian equation is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
The vector equation is
\(\vec{r}=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})\)

Question 6.
Find the cartesian equation of the line which passes through the point (- 2, 4, -5) and parallel to the line given by \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
Solution:
The cartesian equation of a line passing through the point (x₁, y₁, z₁) and parallel to the line with direction ratios a, b, c is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
Hence the cartesian equation of the required line is \(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)

Question 7.
The cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\) write its vector form.
Solution:
The equation of the line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\)
The required line passes through the point (5, – 4, 6) and is parallel to the vector \(3 \hat{i}+7 \hat{j}+2 \hat{k}\). Let \(\vec{r}\) be the position vector of any point on the line, then the vector equation of the line is
\(\vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})\)

Question 8.
Find the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3).
Solution:
Let \(\vec{a}\) and \(\vec{b}\) be the position vectors of the point A (0, 0,0) and B (5, – 2, 3)

Question 9.
Find the vector and cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).
Solution:
Let \(\vec{a}\) and \(\vec{b}\) be the position vector of the point A (3, -2, -5) and B (3, -2, 6)
\(\vec{a}=3 \hat{i}-2 \hat{j}-5 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+6 \hat{k}\)
∴ \(\vec{b}-\vec{a}=0 \hat{i}+0 \hat{j}+11 \hat{k}\)
Let \(\vec{r}\) be the position vector of any point on the line. Then the vector equation of the line is \(\vec{r}\) = \(\vec{a}\) + λ(\(\vec{b}\) – \(\vec{a}\))
\(\vec{r}\) = \((3 \hat{i}-2 \hat{j}-5 \hat{k})+\lambda(11 \hat{k})\)
The cartesian equation is
\(\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}\)

Question 10.
Find the angle between the following pair of lines
(i) \(\overrightarrow { r } =2\hat { i } -5\hat { j } +\hat { k } +\lambda (3\hat { i } +2\hat { j } +6\hat { k } )\)
\(and\quad \overrightarrow { r } =7\hat { i } -6\hat { j } +\mu (\hat { i } +2\hat { j } +2\hat { k } )\)
(ii) \(\overrightarrow { r } =3\hat { i } +\hat { j } -2\hat { k } +\lambda (\hat { i } -\hat { j } -2\hat { k } )\)
\(\overrightarrow { r } =2\hat { i } -\hat { j } -56\hat { k } +\mu (3\hat { i } -5\hat { j } -4\hat { k } )\)
Solution:

Question 11.
Find the angle between the following pair of lines
(i) \(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\)
(ii) \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1} \text { and } \frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
Solution:
i. The direction ratios of the first line are 2, 5, – 3 and the direction ratios of the second line are -1, 8,4
Let θ be the acute angle between the lines, then

ii. The direction ratios of the first line are 2, 2, 1 and the direction ratios of the second line are 4, 1, 8 Let θ be the acute angle between the lines.

Question 12.
Find the values of p so that the lines
\(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles
Solution:
The given lines are

Question 13.
Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) are perpendicular to each other
Solution:
The direction ratios of the first line are 7, – 5, 1 and the direction ratios of second line are 1, 2, 3
a₁a₂ + b₁b₂ + c₁c₂ = (7)(1) + (-5)(2) + (1)(3) = 7 – 10 + 3 = 0
∴ The lines are perpendicular to each other.

Question 14.
Find the shortest distance between the lines
\(\overrightarrow { r } =(\hat { i } +2\hat { j } +\hat { k } )+\lambda (\hat { i } -\hat { j } +\hat { k } )\) and \(\overrightarrow { r } =(2\hat { i } -\hat { j } -\hat { k } )+\mu (2\hat { i } +\hat { j } +2\hat { k } )\)
Solution:
\(\overrightarrow { r } =(\hat { i } +2\hat { j } +\hat { k } )+\lambda (\hat { i } -\hat { j } +\hat { k } )\) … (1)
\(\overrightarrow { r } =(2\hat { i } -\hat { j } -\hat { k } )+\mu (2\hat { i } +\hat { j } +2\hat { k } )\) … (2)

Question 15.
Find the shortest distance between the lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Solution:
Shortest distance between the lines

Question 16.
Find the distance between die lines whose vector equations are:
\(\overrightarrow { r } =(\hat { i } +2\hat { j } +3\hat { k) } +\lambda (\hat { i } -3\hat { j } +2\hat { k } )\) and \(\overrightarrow { r } =(4\hat { i } +5\hat { j } +6\hat { k) } +\mu (2\hat { i } +3\hat { j } +\hat { k } )\)
Solution:

Question 17.
Find the shortest distance between the lines whose vector equations are
\(\overrightarrow { r } =(1-t)\hat { i } +(t-2)\hat { j } +(3-2t)\hat { k }\) and \(\overrightarrow { r } =(s+1)\hat { i } +(2s-1)\hat { j } -(2s+1)\hat { k }\)
Solution:
The given equation can be reduced as
\(\vec{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+t(-\hat{i}+\hat{j}-2 \hat{k})\)
and \(\vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2 \hat{j}-2 \hat{k})\)
Comparing with the standard equation, we get

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NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.1

Question 1.
If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
Solution:
Let the direction angles be α, ß, γ.
i.e., α = 90°, ß = 135° and γ = 45°
The direction cosines are cos α, cos ß, cos γ = cos 90°, cos 135°, cos45°
= 0, \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\)

Question 2.
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Solution:
Let the direction angles be α, ß, γ. Angles of the line. Since this makes equal angles with the axes, we get

Question 3.
If a line has the direction ratios – 18, 12, – 4 then what are its direction cosines?
Solution:
The direction ratios are – 18, 12, – 4
\(\sqrt{(-18)^{2}+(12)^{2}+(-4)^{2}}=\sqrt{384}\) = 22
The direction cosines are
\(\frac{-18}{22}, \frac{12}{22}, \frac{-4}{22}=\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\)

Question 4.
Show that the points (2, 3, 4) (- 1, – 2, 1), (5, 8, 7) are collinear.
Solution:
Let A (2, 3, 4), B (- 1, – 2, 1) and C (5, 8, 7) be the points.
Direction ratios of AB
= – 1 – 2, – 2 – 3, 1 – 4
= – 3, – 5, – 3
The direction ratios of BC
= 5 + 1, 8 + 2, 7 – 1 = 6, 10, 6
The direction ratios of AB and BC are proportional
∴ AB and BC are parallel.
Hence A, B, C are collinear.

Question 5.
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (- 1, 1, 2) and (- 5, – 5,- 2).
Solution:
Let A (3, 5, – 4) B (- 1, 1, 2) and C (- 5, – 5, – 2) are the vertices of ∆ABC

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Exercise 10.4

Class 12 Maths Ex 10.4 Question 1.
Find \(|\vec{a} \times \vec{b}|\), if \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\).
Solution:

Question 2.
Find a unit vector perpendicular to each of the vector \(\overrightarrow { a } +\overrightarrow { b } \quad and\quad \overrightarrow { a } -\overrightarrow { b } \), where \(\overrightarrow { a } =3\hat { i } +2\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =\hat { i } +2\hat { j } -2\hat { k } \)
Solution:

Question 3.
If a unit vector \(\vec{a}\) makes angles \(\frac { π }{ 3 }\) with \(\hat{i}\), \(\frac { π }{ 4 }\) with \(\hat{j}\) and an acute angle θ with k, then find θ and hence, the components of \(\vec{a}\).
Solution:
The direction cosines of \(\vec{a}\) are

Since \(\vec{a}\) is a unit vector, its components are the direction cosines

Question 4.
Show that
\((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)
Solution:
\((\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})=\vec{a} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{a}-\vec{b} \times \vec{b}\)
= \(\overrightarrow{0}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}-\overrightarrow{0}\)
= 2(\(\vec{a}\) x \(\vec{b}\))

Question 5.
Find λ and μ if
\(\left( 2\hat { i } +6\hat { j } +27\hat { k } \right) \times \left( \hat { i } +\lambda \hat { j } +\mu \hat { k } \right)\) = \(\vec{0}\)
Solution:
\((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\)
\(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|=\overrightarrow{0}\)
\(\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=\overrightarrow{0}\)
Equating the corresponding components, we get
6µ – 27λ = 0 … (1)
– 2µ + 27 = 0 … (2)
2λ – 6 = 0 … (3)
(2) → µ = \(\frac { 27 }{ 2 }\), (3) → λ = 3
Substituting the values of λ and µ in (1),
we get 6(\(\frac { 27 }{ 2 }\)) – 27(3) = 81 – 81 = 0
(1) satisfy λ = 3 and µ = \(\frac { 27 }{ 2 }\)
Hence λ = 3, µ = \(\frac { 27 }{ 2 }\)

Question 6.
Given that \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) x \(\vec{b}\) = \(\vec{0}\) What can you conclude about the vectors \(\vec{a}\) and \(\vec{b}\)?
Solution:
\(\vec{a}\).\(\vec{b}\) = 0 ⇒ \(\vec{a}\) = \(\vec{0}\) or \(\vec{b}\) = \(\vec{0}\) or \(\vec{a}\) ⊥ \(\vec{b}\)
\(\vec{a}\) x \(\vec{b}\) = \(\vec{0}\) ⇒ \(\vec{a}\) = \(\vec{0}\) or \(\vec{b}\) = \(\vec{0}\) or \(\vec{a}\) || \(\vec{b}\)
Since \(\vec{a}\).\(\vec{b}\) = 0 and \(\vec{a}\) x \(\vec{b}\) = 0,
then either \(\vec{a}\) = \(\vec{0}\) or \(\vec{b}\).\(\vec{0}\)
\(\vec{a}\) ⊥ \(\vec{b}\) and \(\vec{a}\)||\(\vec{b}\) are not possible at the same time.

Question 7.
Let the vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) be given as \({ a }_{ 1 }\hat { i } +{ a }_{ 2 }\hat { j } +{ a }_{ 3 }\hat { k } ,{ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } ,{ c }_{ 1 }\hat { i } +{ c }_{ 2 }\hat { j } +{ c }_{ 3 }\hat { k }\), then show that \(\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\).
Solution:

Question 8.
If either \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text {, then } \vec{a} \times \vec{b}=\overrightarrow{0}\). Is the converse true? Justify your answer with an example.
Solution:
\(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}\)
If \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text {, then } \vec{a} \times \vec{b}=\overrightarrow{0}\).
The converse need not be true.
For example, consider the non-zero parallel

Question 9.
Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).
Solution:

Question 10.
Find the area of the parallelogram whose adjacent sides are determined by the vectors \(\overrightarrow { a } =\hat { i } -\hat { j } +3\hat { k } ,\overrightarrow { b } =2\hat { i } -7\hat { j } +\hat { k } \)
Solution:

Question 11.
Let the vectors \(\vec{a}\) and \(\vec{b}\) such that |\(\vec{a}\)| = 3 and |\(\vec{b}\)| = \(\frac{\sqrt{2}}{3}\), then \(\vec{a}\) x \(\vec{b}\) is a unit vector if the angle between \(\vec{a}\) and \(\vec{a}\) is
(a) \(\frac { \pi }{ 6 } \)
(b) \(\frac { \pi }{ 4 } \)
(c) \(\frac { \pi }{ 3 } \)
(d) \(\frac { \pi }{ 2 } \)
Solution:
Let θ be the angle between vectors \(\vec{a}\) and \(\vec{b}\).
Since \(\vec{a}\) x \(\vec{b}\) is a unit vector,

Question 12.
Area of a rectangles having vertices
\(\left( -\hat { i } +\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right), \left( \hat { i } +\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right)\)
\(\left( \hat { i } -\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right), \left( -\hat { i } -\frac { 1 }{ 2 } \hat { j } +4\hat { k } \right)\)
(a) \(\frac { 1 }{ 2 }\) sq units
(b) 1 sq.units
(c) 2 sq.units
(d) 4 sq.units
Solution:

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NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.3

Ex 11.3 Class 12 NCERT Solutions Question 1.
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2
(b) x + y + z = 1
(c) 2x + 3y – z = 5
(d) 5y + 8 = 0
Solution:
a. The equation can be written as 0x + 0y + 1z = 2,
which is in the normal form Hence the direction cosines of the normal to the plane are 0, 0, 1 and distance from the orgin = 2

b. Equation of the plane is x + y + z = 1 … (1)
The direction ratios of the normal to the plane are 1, 1, 1
∴ Direction cosines are
\(\frac{1}{\sqrt{1+1+1}}, \frac{1}{\sqrt{1+1+1}}, \frac{1}{\sqrt{1+1+1}}\)
\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
Dividing (1) by \(\sqrt{3}\), we get
\(\frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}}=\frac{1}{\sqrt{3}}\)
This is of the form lx + my + nx = d where d is the distance from the orgin.
∴ Distance from the orgin = \(\frac{1}{\sqrt{3}}\)

c. The direction ratios of the normal to the plane are 2, 3, -1
The direction cosines are

d. The direction ratios of the normal to the plane are 0, 5, 0

Exercise 11.3 Class 12 NCERT Solutions Question 2.
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \(3\hat { i } +5\hat { j } -6\hat { k }\)
Solution:

Question 3.
Find the Cartesian equation of the following planes.
(a) \(\overrightarrow { r } \cdot (\hat { i } +\hat { j } -\hat { k) }\) = 2
(b) \(\overrightarrow { r } \cdot (\hat { 2i } +3\hat { j } -4\hat { k) }\) = 1
(c) \(\overrightarrow { r } \cdot [(s-2t)\hat { i } +(3-t)\hat { j } +(2s+t)\hat { k) }\) = 15
Solution:
Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) be the position vector of any point on the plane.
\(\overrightarrow { r } \cdot (\hat { i } +\hat { j } -\hat { k) }\) = 2
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})\)
x + y – z = 2 is the cartesian equation.

b. Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ \(\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})\) = 1,
\((x \hat{i}+\hat{y j}+z \hat{k}) \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})\) = 1
2x + 3y – 4z = 1 is the cartesian equation.

c. Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
\(\vec{r} \cdot((s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k})\) = 15
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot((s-2 t) \hat{i}+(3-t) \hat{j}\) + \((2 s+t) \hat{k})\) = 15
(s – 2t)x + (3 – t)y + (2s + t)z = 15, is the cartesian equation.

Question 4.
In the following cases find the coordinates of the foot of perpendicular drawn from the origin
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Solution:
Let A be the foot of the perpendicular drawn from the orgin O.
Equation of the plane is 2x + 3y + 4z – 12 = 0
∴ The d.r’s of the normal to the plane are 2, 3, 4
Equation of the line OA is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
\(\frac{x-0}{2}=\frac{y-0}{3}=\frac{z-0}{4}\)
i.e. \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\lambda\)
∴ Any point on the line OA is (2λ, 3λ, 4λ) Let it be A.
A satisfies the equation 2x + 3y + 4z – 12 = 0
4λ + 9λ + 16λ – 12 = 0
29λ – 12 = 0

Another Method:
The equation of the plane is 2x + 3y + 42 = 12 … (1)
Dividing (1) by \(\), we get
\(\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29}\) is of the form lx + my + nz = d
The direction cosines of the normal to the plane are l = \(\frac{2}{\sqrt{29}}\), m = \(\frac{3}{\sqrt{29}}\), n = \(\frac{4}{\sqrt{29}}\)
and the distance from the origin to the plane d = \(\frac{12}{\sqrt{29}}\)
Foot of the perpendicular = (Id, md, nd)
= \(\left(\frac{2}{\sqrt{29}}\left(\frac{12}{\sqrt{29}}\right), \frac{3}{\sqrt{29}}\left(\frac{12}{\sqrt{29}}\right), \frac{4}{\sqrt{29}}\left(\frac{12}{\sqrt{29}}\right)\right)\)
= \(\left(\frac{24}{29}, \frac{36}{29}, \frac{48}{29}\right)\)

b. The equation of the plane is , 3y + 4z = 6 … (1)
Dividing (1) by \(\sqrt{3^{2}+4^{2}}\) = 5, we get
\(\frac{3}{5} y+\frac{4}{5} z=\frac{6}{5}\), is in the form lx + my + nz = d
∴ l = 0, m = \(\frac { 3 }{ 5 }\), n = \(\frac { 4 }{ 5 }\), d = \(\frac { 6 }{ 5 }\)
The foot of the perpendicular from the origin = (Id, md, nd)
= \(\left(0\left(\frac{6}{5}\right), \frac{3}{5}\left(\frac{6}{5}\right), \frac{4}{5}\left(\frac{6}{5}\right)\right)\)
= \(\left(0, \frac{18}{25}, \frac{24}{25}\right)\)

c. The equation of the plane is
x + y + z = 1 … (1)

d. Equation of the plane is
5y = – 8 or – 5y = 8
Dividing (1) by \(\sqrt{(-5)^{2}}\) = 5, we get
\(\frac{-5}{5} y=\frac{8}{5} \quad \text { or } \quad-y=\frac{8}{5}\), is in the form
lx + my + nz = d
l = 0, m = – 1, n = 0, d = \(\frac { 8 }{ 5 }\)
The foot of the perpendicular from the origin = (Id, md, nd)
= \(\left(0\left(\frac{8}{5}\right),-1\left(\frac{8}{5}\right), 0\left(\frac{8}{5}\right)\right)\)
= \(\left(0, \frac{-8}{5}, 0\right)\)

Question 5.
Find the vector and cartesian equation of the planes
(a) that passes through the point (1,0, -2) and the normal to the plane is \(\hat { i } +\hat { j } -\hat { k }\)
(b) that passes through the point (1,4,6) and the normal vector to the plane is \(\hat { i } -2\hat { j } +\hat { k }\)
Solution:
(a) The plane passes through the point (1, 0, – 2)
∴ \(\vec{a}=\hat{i}+0 \hat{j}-2 \hat{k}\)
The normal vector to the plane \(\vec{n}=\hat{i}+\hat{j}-\hat{k}\)
The vector equation of the plane passing through \(\vec{a}\) and perpendicular to \(\vec{n}\) is (r – a).n = 0

The cartesian equation is x + y – z – 3 = 0

(b) The plane passes through (1, 4, 6) and the direction ratios of the normal to the plane are 1, -2, 1.
The cartesian equation of the plane passing through (x₁, T₁, Z₁) and perpendicular to the line with direction ratios a, b, c is
a(x – x₁) + b(y – y₁) + c(z – z₁) = 0
Hence the cartesian equation of the plane is 1 (x – 1) – 2(y – 4) + 1(z – 6) = 0
x – 2y + z + 1 = 0
The vector equation of the plane is
\(\vec{r} \cdot(\hat{i}-2 \hat{j}+\hat{k})+1=0\)

Question 6.
Find the equations of the planes that passes through three points
(a) (1, 1, – 1) (6, 4, – 5), (- 4, -2, 3)
(b) (1, 1, 0), (1, 2, 1), (-2, 2, -1)
Solution:
Let the equation of plane passing through (1, 1, – 1) be
a(x – 1) + b(y – 1) + c(z + 1 ) = 0 … (1)
Since (6, 4, – 5) is a point on (1), we get
a(6 – 1) + b(4 – 1) + c(- 5 + 1) = 0
i.e., 5a + 3b – 4c = 0 … (2)
Since (- 4, – 2, 3) is a point on (1), we get
a(- 4 – 1) + b(- 2 – 1) + c(3 + 1) = 0
– 5a – 3b + 4c = 0
5a + 3b – 4c = 0 … (3)
From (2) and (3), by the rule of cross multiplication, we get
\(\frac{a}{-12+12}=\frac{b}{-20+20}=\frac{c}{15-15}\)
\(\frac{a}{0}=\frac{b}{0}=\frac{c}{0}\)
a = 0, b = 0 and c = 0
Hence, there is no unique plane passing through the given points.

Another Method:
The plane passes through
(1, 1, – 1),(6, 4, – 5), (- 4, – 2, 3)
The equation of the plane passing through
\(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right) \text { and }\left(x_{3}, y_{3}, z_{3}\right)\) is

R₂ and R₃ are proportional.
Hence we get 0 = 0
So we cannot find the equation of the plane passing through these points in a unique way.

Question 7.
Find the intercepts cut off by the plane 2x+y-z = 5.
Solution:
Equation of the plane is 2x + y- z = 5 x y z
Dividing by 5:
\(\Rightarrow \frac { x }{ \frac { 5 }{ 2 } } +\frac { y }{ 5 } -\frac { z }{ -5 }\) = 1
∴ The intercepts on the axes OX, OY, OZ are \(\frac { 5 }{ 2 }\), 5, – 5 respectively.

Question 8.
Find the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane.
Solution:
Any plane parallel to ZOX plane is y=b where b is the intercept on y-axis.
∴ b = 3.
Hence equation of the required plane is y = 3.

Question 9.
Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2,2,1).
Solution:
Given planes are:
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0
Any plane through their intersection is
3x – y + 2z – 4 + λ(x + y + z – 2) = 0
point (2, 2, 1) lies on it,
∴ 3 x 2 – 2 + 2 x 1 – 4 + λ(2 + 2 + 1 – 2) = 0
⇒ λ = \(\frac { -2 }{ 3 }\)
Now required equation is 7x – 5y + 4z – 8 = 0

Question 10.
Find the vector equation of the plane passing through the intersection of the planes \(\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =7,\overrightarrow { r } \cdot \left( 2\hat { i } +5\hat { j } +3\hat { k } \right) =9\) and through the point (2, 1, 3).
Solution:
Here \(\vec{n}_{1}=2 \hat{i}+2 \hat{j}-3 \hat{k}, d_{1}=7\)
\(\vec{n}_{2}=2 \hat{i}+5 \hat{j}+3 \hat{k}, d_{2}=9\)
The required vector equation is

is the required vector equation of the plane.

Another method:
The equation of the planes are
2x + 2y – 3z = 7 … (1)
2x + 5y + 3z = 9 … (2)
The equation of the plane through the inter-section of (1) and (2) is (2x + 2y – 3z) + λ(2x + 5y + 3z) = 7 + 9λ … (3)
Since the plane (3) passes through (2, 1, 3), we get
(2(2) + 2(1) – 3(3)) + λ (2(2) + 5(1) + 3(3)) = 7 + 9λ
– 3 + 18λ = 7 + 9λ
9 λ = 10 ∴ λ = \(\frac { 10 }{ 9 }\)
(3) gives the cartesian equation of the plane.
(3) → (2x + 2y – 3z) + \(\frac { 10 }{ 9 }\) (2x + 5y + 3z)
= 7 + 9(\(\frac { 10 }{ 9 }\))
Multiplying by 9, we get 18x + 18y – 27z + 20x + 50y + 30z = 63 + 90 38x + 68y + 3z = 153
The vector equation is
\(\bar{r} \cdot(38 \hat{i}+68 \hat{j}+3 \hat{k})\) = 153

Question 11.
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = o.
Solution:
The equation of the required plane is (x + y + z – 1) + λ(2x + 3y + 4z – 5) = 0

Question 12.
Find the angle between the planes whose vector equations are \(\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =5,\overrightarrow { r } \cdot \left( 3\hat { i } -3\hat { j } +5\hat { k } \right)\) = 3
Solution:
The angle θ between the given planes is

Question 13.
In the following determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angle between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0.
Solution:
a. Comparing with the general equation, we get
a₁ = 7 b₁ = 5 c₁ = 6 and
a₂ = 3 b₂ = – 1 c₂ = – 10
a₁a₂ + b₁b₂ + c₁c₂ = 7(3) + 5(- 1) + 6(- 10) ≠ 0
∴ The lines are not perpendicular

b. a₁ = 2 b₁ = 1 c₁ = 3 and
a₂ = 1 b₂ = – 2 c₂ = 0
a₁a₂ + b₁b₂ + c₁c₂ = 2 – 2 + 0 = 0
∴ The two lines are perpendicular.

c. a₁ = 2 b₁ = – 2 c₁ = 4 and
a₂ = 3 b₂ = – 3 c₂ = 6
\(\frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{2}{3}, \frac{c_{1}}{c_{2}}=\frac{4}{6}=\frac{2}{3}\)
i.e., \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
∴ The two lines are parallel.

d. a₁ = 2 b₁ = – 1 c₁ = 3 and
a₂ = 2 b₂ = – 1 c₂ = 3
\(\frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=1, \frac{c_{1}}{c_{2}}=1\)
∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
∴ The two lines are parallel.

e. a₁ = 4 b₁ = 8 c₁ = 1 and
a₂ = 0 b₂ = 1 c₂ = 1
a₁a₂ + b₁b₂ + c₁c₂ = 2(3) + – 2(3) + 4(6) ≠ 0
∴ The two lines are perpendicular.

Question 14.
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point           Plane
(a) (0, 0,0) 3x – 4y + 12z = 3
(b) (3,-2,1) 2x – y + 2z + 3 = 0.
(c) (2,3,-5) x + 2y – 2z = 9
(d) (-6,0,0) 2x – 3y + 6z – 2 = 0
Solution:
(a) Given plane: 3x – 4y + 12z – 3 = 0
∴ The distance from the point (0, 0, 0) to

b. The equation of the plane is
2x – y + 2z + 3 = 0
∴ The distance from the point (3, – 2, 1) to the given plane
= \(\left|\frac{2(3)-1(-2)+2(1)+3}{\sqrt{4+1+4}}\right|=\frac{13}{3}\)

c. The equation of the plane is x + 2y – 2z – 9 = 0
a = 1, b = 2, c = – 2, d = 9
∴ The distance from the point (2, 3, – 5) to the plane
= \(\left|\frac{a x_{1}+b y_{1}+c z_{1}-d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|=\left|\frac{2+2(3)-2(-5)-9}{\sqrt{1+4+4}}\right|\)
= \(\frac{9}{3}\) = 3

d. The equation of the plane is 2x – 3y + 6z – 2 = 0
∴ The perpendicular distance from the point (- 6, 0, 0) to the plane
= \(\left|\frac{2(-6)-3(0)+6(0)-2}{\sqrt{4+9+36}}\right|\)
= \(\left|\frac{14}{\sqrt{49}}\right|=\frac{14}{7}\) = 2