CBSE Class 12

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

These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-11-ex-11-1/

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

These NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra Ex 10.4 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-10-ex-10-4/

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:

These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-11-ex-11-3/

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

These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-12-ex-12-1/

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.1

Ex 12.1 Class 12 Question 1.
Maximize Z = 3x + 4y
subject to the constraints:
x + y ≤ 4, x ≥ 0, y ≥ 0.
Solution:
The objective function is Z = 3x + 4y
The constraints are x + y ≤ 4, x ≥ 0, y ≥ 0.

∴ Maximum value of Z is 16 at B (0, 4).

Question 2.
Minimize Z = – 3x + 4y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Solution:
The objective function is Z = – 3x + 4y The constraints are x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0

∴ Maximum value of Z is – 12 at A (4, 0).

Question 3.
Maximize Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
Solution:
The objective function is Z = 5x + 3y
The constraints are 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
The feasible region is shaded in the figure. The point B is obtained by solving the equation 5x + 2y = 10 and 3x + 5y = 15.

Question 4.
Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Solution:
The objective function is Z = 3x + 5y
The constraints are x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
The feasible region is shaded in the figure.

From the table the minimum value of Z is 7. Since the feasible region is unbounded, 7 may or may not be the minimum value of Z. Consider the graph of the inequality 3x + 5y < 7. This half plane has no point in common with the feasible region. Hence the minimum value of Z is 7 at B (\(\frac { 3 }{ 2 }\), \(\frac { 1 }{ 2 }\))

Question 5.
Maximize Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
Solution:

The objective function is Z = 3x + 5y
The constraints are x + 2y < 10,
3x + y ≤ 15, x, y ≥ 0.
The feasible region is shaded in the figure. We use the corner point method to find the maximum of Z
Maximum value of Z is 18 at B(4, 3)

Question 6.
Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Solution:
Consider 2x + y ≥ 3
Let 2x + y = 3
⇒ y = 3 – 2x

(0,0) is not contained in the required half plane as (0, 0) does not satisfy the in equation 2x + y ≥ 3.
Again consider x + 2y ≥ 6
Let x + 2y = 6
⇒ \(\frac { x }{ 6 } +\frac { y }{ 3 }\) = 1
Here also (0,0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its comers are A (6,0) and B (0,3). At A, Z = 6 + 0 = 6
At B, Z = 0 + 2 × 3 = 6
We see that at both points the value of Z = 6 which is minimum. In fact at every point on the line AB makes Z = 6 which is also minimum.

Question 7.
Minimise and Maximise Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0
Solution:
The objective function is Z = 5x + 10v.
The constraints are x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0

The feasible region is shaded in the figure. We use the comer point method to find the maximum/minimum value of Z.

Minimum value of Z is 300 at A(60,0).
The maximum value of Z is 600 at all points on the line joining B(120, 0) and C(60, 30).

Question 8.
Minimize and maximize Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.
Solution:
The objective function is Z = x + 2y
The constraints are x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0

The feasible region is shaded in the figure. We use corner point method to find maxi-mum/minimum value of Z.

Maximum value of Z is 400 at B(0,200). The minimum value of Z is 100 at all points of the line joining the points (0,50) and (20,40).

Question 9.
Maximize Z = – x + 2y, subject to the constraints: x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
Solution:

The objective function is Z = – x + 2y
The constraints are x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
The feasible region is shaded in the figure.

From the table the maximum value of Z is 1. Since the feasible region is unbounded 1 may or may not be the maximum value of Z
Consider the graph of the inequality – x + 2y > 1. This half plane has points common with the feasible region. Hence there is no maximum value for Z.

Question 10.
Maximize Z = x + y subject to x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0
Solution:
The objective function is Z = x + y
The constraints are x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0

There is no point satisfying the constraints simultaneously. Thus the problem has no feasible region. Hence no maximum value for Z.

These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-12-ex-12-2/

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Exercise 12.2

Ex 12.2 Class 12 NCERT Solutions Question 1.
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/ kg of Vitamin A and 5 units/ kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Solution:
Let the mixture contain x units of food P and y units of food Q. We have the data as

Vitamin A constaint: 3x + Ay ≥ 8
Vitamin B constraint: 5x + 2y ≥ 11
Cost function : Z = 60x + 80y
The L.P.P is Minimise Z = 60x + 80y
subject to the constraints 3x + 4y ≥8, 5x + 2y ≥ 11, x, y ≥ 0.

The feasible region is shaded in the figure and is unbounded.
We use comer point method to find the mini-mum of Z
From the table, the minimum value of Z is 160. Since the feasible region is unbounded j 160 may or may not be the minimum value i of Z. Consider the graph of the inequality 60x + 80y ≤ 160
i.e., 3x + 4y ≤ 8
This half plane has no point in common with the feasible region. Then the minimum value of Z is 160 at the points on the line segment joining A\(\frac { 8 }{ 3 }\), 0) and B(2, \(\frac { 1 }{ 2 }\))

Exercise 12.2 Class 12 Maths Ncert Solutions In Hindi Question 2.
One kind of cake requires 200 g of flour and 25g of fat, and another kind of cake requires 100 g of flour and 50 g of fat Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients, used in making the cakes.
Solution:
Let x be the number of cakes of type I and y be the number of cakes of type II.
We make use of the following table to write the L.P.P.
We have the data as

Flour constraint: 200x + 100y ≤ 5000
i.e., 2x + y ≤ 50
Fat constraint: 25x + 50y ≤ 1000
i.e., x + 2y ≤ 40
Total number of cakes : Z = x + y
The L.P.P is Maximise Z = x + y
subject to the constraints
2x + y ≤ 50, x + 2y ≤ 40, x, y >0.

∴ Maximum value of Z is 30 at B(20, 10).
Hence maximum number of cakes is 30, of which 20 are of type 1 and 10 are of type II.

Chapter 12 Exercise 12.2 NCERT Solutions Question 3.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.
i. What number of rackets and bats must be made if the factory is to work at full capacity?
ii. If the profit on a racket and on a bat is ₹ 20 and ₹ 10 respectively, find the maximum profit of the factory when it works at full capacity.
Solution:
i. Let x be the number of tennis rackets and that y be the cricket bats.

Machine constraint: 1.5x + 3y ≤ 42
i.e., x + 2y ≤ 28
Craftman’s constraint: 3x + y ≤ 24
Profit function : Z = 20x + 10y
The L.P.P is Maximise Z = 20x + 10y subject to the constraints
x + 2y ≤ 28, 3x + y ≤ 24, x, y ≥ 0

∴ Maximum value of Z is 200 at B(4, 12).
Hence 4 tennis rackets and 12 cricket bats can be made if the factory is to work at full capacity.

ii. The maximum profit is ₹ 200/-

Ncert Solutions For Class 12 Maths Chapter 12 Question 4.
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ₹ 17.50 per package on nuts and ₹ 7.00 per package on bolts. How many package of each should be produced each day so as to maximise his profit, if he operates his machine for at the most 12 hours a day?
Solution:
Let x and y be the number of packages of nuts and bolts.
∴ We have the data as

Machine A constraint: x + 3y ≤ 12
Machine B constraint: 3x + y ≤ 12
Profit function : Z = 17.5x + 7y
The L.P.P is maximise Z = 17.5x + 7y subject to the constraints 1x + 3y ≤ 12, 3x+ 1y ≤ 12, x, y > 0

Maximum value of Z is 73.50 at B(3, 3)
Hence the manufacturer must produce 3 packages of nuts and 3 packages of bolts to maximise his profit. The maximum profit is ₹ 73.50.

Maths Chapter 12 Exercise 12.2 Question 5.
A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of ₹ 7 and screws B at a profit of ₹ 10. Assuming that he can sell all the screws he manufactures, how many pack¬ages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
Solution:
Let x be the number of packages of screw A and y be that of screw B.
We have the data as

Automatic machine constraint: 4x + 6y ≤ 240
Hand operated machine constraint: 6x + 3y ≤ 240
Profit function : Z = 7x + 10y
The L.P.P is maximise Z = 1x + 10y subject to the constraints 4x + 6y ≤ 240, 6x + 3y ≤ 240, x, y ≥ 0.

Maximum value of Z is 410 at B(30, 20).
Hence the factory should manufacture 30 packages of screw A and 20 packages of screw B to maximise the profit. The maxi-mum profit is ₹ 410.

Class 12 Chapter 12 Maths Ncert Solutions Question 6.
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting ma-chine for at the most 12 hours. The profit from the sale of a lamp is ₹ 5 and that from a shade is ₹ 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
Solution:
Let the manufacturer produce x lamps and v wooden shades.

Grinding/cutting constraint: 2x + y ≤ 12
Sprayer constraint: 3x + 2y ≤ 20
Profit function : Z = 5x + 3y
∴ The L.P.P is maximise Z = 5x + 3y subject to the constraints 2x + y ≤ 12, 3x + 2y ≤ 20, x, y ≥ 0.

The maximum value of Z is 32 at B(4, 4). Hence to get maximum profit 4 pedestal lamps and 4 wooden shades are to be manufactured. The maximum profit obtained is ₹ 32.

Exercise 12.2 NCERT Solutions Question 7.
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should be company manufacture in order to maximise the profit?
Solution:
Let x be the number of souvenirs of type A and y be the number of souvenirs of type B to be manufactured.
We have the data as

Cutting constraint: 5x+ 8y ≤ 200
Assembling constraint: 10x + 8y ≤ 240
Profit function : Z = 5x + 6y
Then the L.P.P is maximise Z = 5x + 6y subject to the constraints 5x + 8y ≤ 200, 10x + 8y ≤ 240, x, y ≥ 0.

The maximum value of Z is 160 at B(8,20).
Hence for maximum profit 8 souvenirs of type A and 20 souvenirs of type B should be manufactured.
The maximum profit is ₹ 160.

Exercise 12.2 Class 12 NCERT Solutions Question 8.
A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost ₹ 25000 and ₹ 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ₹ 70 lakhs and if his profit on the desktop model is ₹ 4500 and on portable model is ₹ 5000.
Solution:
Let the merchant plan to stock x desktop computers and y portable models.
We have the data as

Demand constraint: x + y ≤ 250
Cost constraint:
25000 x + 40000 y ≤ 7000000
i.e., 5x + 8 ≤ 1400
Profit function : Z = 4500x + 5000y
The L.P.P is maximise Z = 4500x + 5000y
subject to the constraints x + y ≤ 250, 5x + 8y ≤ 1400, x, y ≥ 0.

Maximum value of Z is 1150000 at B(200,50) Hence 200 desktop models and 50 portable models are to be stored to get maximum profit.
The maximum profit is ₹ 1150000.

12.2 Class 12 NCERT Solutions Question 9.
A diet is to contain atlest 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs ₹ 4 per unit food and F₂ costs ₹ 6 per unit. One unit of food F₁ contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Solution:
Let there be x units of food F₁ and y units of food F₂.

Vitamin A constraint: 3x + 6v ≥ 80
Minarals constraint: 4x + 3y ≥ 100
Cost function is : Z = 4x + 6y
The L.P.P is minimise Z = 4x + 6y subject to the constraints 3x + 6v ≥ 80,
4x + 3y ≥ 100, x,y ≥0.

From the table the minimum value of Z is 104. Since the region is unbounded, 104 may or may not be the minimum value of Z.
Consider the inequality 4x + 6y ≤ 104. This half plane has no point common with the feasible region. Hence the minimum value of Z is 104 at B (24, \(\frac { 4 }{ 3 }\))
∴ Minimum cost for diet is ₹ 104

Question 10.
There are two types of fertilisers F₁ and F₂F₁ consists of 10% nitrogen and 6% phosphoric acid and F₂ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that he needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁, costs ₹ 6/kg and F, costs ₹5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Solution:
Let x kg of fertilizer F₁ and y kg of fertilizer
F₂ be used.
We have the data as

Nitrogen constraint: 0.1 x + 0.05 y ≥ 14
i.e., 2x + y ≥ 280
Phosphoric acid constraint: 0.06x + 0.1 y ≥ 14
3x + 5y ≥ 700
Cost function : Z = 6x +5y
The L.P.P is minimise Z = 6x + 5y subject to the constraints 2x + y ≥ 280 and 3x + 5y ≥ 700, x ≥ 0, y ≥ 0

From the table the minimum value of Z is 1000. Since the region is unbounded 1000 may or may not be the minimum value of Z.
Consider the inequality 6x + 5y ≤ 1000
This half plane has no point common with the feasible region
∴ Minimum value of Z is 1000 at B( 100,80).
Hence 100 kg of fertilizer F₁ and 80 kg of fertilizer F₂ is used. The minimum cost of these fertilizers is ₹ 1000.

Question 11.
The corner points of the feasible region de-termined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where р, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0,5) is
a. p = q
b. p = 2q
с. p = 3q
d. q – 3p
Solution:
d. q – 3p
Since maximum Z occurs at (3, 4) and (0, 5), we get Z = 3p + 4q and Z = 0 + 5q
Since both values are same.
∴ 3p + 4q = 5q Hence 3p = q