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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. \(\frac { π }{ 4 }\)
