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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 1.
Find the equation of the plane passing through the line of intersection of the planes \(\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})\) = 3 and 3x – 5y + 4z + 11 = 0, and the point (-2, 1, 3).
Solution:
Given planes are
\(\vec{r} \cdot(2 \hat{i}-7 \hat{j}+4 \hat{k})\) = 3
2x – 7y + 4z – 3 = 0 and 3x – 5y + 4z + 11 = 0
Equation of a plane which passes through the line of intersection of the planes
(2x – 7y + 4z – 3) + λ(3x – 5y + 4z + 11) = 0 …………… (1)
This passes through the point (-2, 1, 3).
(1) ⇒ (-4 – 7 + 12 – 3) + λ(-6 – 5 + 12 + 11) = 0
-2 + λ(12) = 0 ⇒ 12λ = 2
λ = \(\frac{2}{12}\) ⇒ λ = \(\frac{1}{6}\)
The required equation is
(1) ⇒ (2x – 7y + 4z – 3) + \(\frac{1}{6}\) (3x – 5y + 4z + 11) = 0
12x – 42y + 24z – 18 + 3x – 5y + 4z + 11 = 0
15x – 47y + 28z – 7 = 0

Question 2.
Find the equation of the plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z + 11 = 3, and at a distance \(\frac{2}{\sqrt{3}}\) from the point (3, 1, -1).
Solution:
Equation of a plane which passes through the line of intersection of the plane
(x + 2y + 3z – 2) + λ(x – y + z – 3) = 0
(1 + λ)x + (2 – λ)y + (3 + λ)z + (-2 – 3λ) = 0 ………….. (1)
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 1

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 3.
Find the angle between the line \(\vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+t(\hat{i}+2 \hat{j}-2 \hat{k})\) and the plane \(\vec{r} \cdot(6 \hat{i}+3 \hat{j}+2 \hat{k})\) = 8.
Solution:
Angle between the line and a plane
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 2

Question 4.
Find the angle between the planes \(\vec{r} \cdot(\hat{i}+\hat{j}-2 \hat{k})\) = 3 and 2x – 2y + z = 2.
Solution:
Angle between given two planes
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 3

Question 5.
Find the equation of the plane which passes through the point (3, 4, -1)and is parallel to the plane 2x – 3y + 5z + 7 = 0. Also, find the distance between the two planes.
Solution:
The required equation parallel to the plane
2x – 3y + 5z + 7 = 0 ………….. (1)
2x – 3y + 5z + λ = 0 ………….. (2)
This passes through (3, 4, -1)
(2) ⇒ 2(3) – 3(4) + 5(-1) + λ = 0
6 – 12 – 5 + 1 = 0
λ = 11
(2) ⇒ The required equation is 2x – 3y + 5z + 11 =0 …………… (3)
∴ Now, distance between the above parallel lines (1) and (3)
a = 2, b = -3, c = 5, d1 = 7, d2 = 11
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 4

Question 6.
Find the length of the perpendicular from the point (1, -2, 3) to the plane x – y + z = 5.
Solution:
Perpendicular length from the point (x1, y1, z1) to the plane ax + by + cz + d = 0 is
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 5

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 7.
Find the point of intersection of the line x – 1 = \(\frac{y}{2}\) = z + 1 with the plane 2x – y + 2z = 2. Also, find the angle between the line and the plane.
Solution:
Any point on the line x – 1 = \(\frac{y}{2}\) = z + 1 is
x – 1 = \(\frac{y}{2}\) = z + 1 = λ,(say)
(λ + 1, 2λ, λ – 1)
This passes through the plane 2x – y + 2z = 2
2(λ + 1) – 2λ + 2(λ – 1) = 2
2λ + 2 – 2λ + 2λ – 2 = 2
λ = 1
∴ The required point of intersection is (2, 2, 0)
Angle between the line and the plane is
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 6

Question 8.
Find the coordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane x + 2y + 3z = 2
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 7
Direction of the normal plane (1, 2, 3)
d.c.s of the PQ is \(\frac{x_{1}-4}{1}=\frac{y_{1}-3}{2}=\frac{z_{1}-2}{3}\) = k
x1 = k + 4, y1 = 2k + 3, z1 = 3k + 2
This passes through the plane x + 2y + 3z = 2
k + 4 + 2(2k + 3) + 3(3k + 2) = 2
k + 4 + 4k + 6 + 9k + 6 = 2
14k = 2 – 16 ⇒ 14k = -14
k = -1
∴ The coordinate of the foot of the perpendicular is (3, 1, -1)
∴ Length of the perpendicular to the plane is
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 8

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 Additional Problems

Question 1.
Find the point of intersection of the line passing through the two points (1, 1, – 1); (-1, 0,1) and the xy-plane.
Solution:
The equation of the line passing through (1, 1,-1) and (-1, 0, 1) is
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 9

Question 2.
Find the co-ordinates of the point where the line \(\vec{r}=(\vec{i}+2 \vec{j}-5 \vec{k})+t(2 \vec{i}-3 \vec{j}+4 \vec{k})\) meets the plane \(\vec{r} \cdot(2 \vec{i}+4 \vec{j}-\vec{k})\) = 3.
Solution:
The equation of the straight line in the cartesian form is
\(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}\) = λ (say)
∴ Any point on this line is of the form (2λ + 1, -3λ, + 2, 4λ – 5)
The cartesian equation of the plane is 2x + 4y – z – 3 = 0
But the required point lies on this plane.
∴ 2(2λ + 1) + 4(-3λ + 2) – (4λ – 5 ) – 3 = 0 ⇒ λ = 1
∴ The required point is (3, -1, -1).

Question 3.
Find the point of intersection of the line \(\vec{r}=(\vec{j}-\vec{k})+s(2 \vec{i}-\vec{j}+\vec{k})\) and xz-plane.
Solution:
The given point is (0, 1, -1); parallel vector is \(2 \vec{i}-\vec{j}+\vec{k}\) .
The equation of the line passing through the point (0, 1, -1) and parallel to the vector \(2 \vec{i}-\vec{j}+\vec{k}\) is
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 10
∴ The required point is (2, 0, 0)

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 4.
Find the meeting point of the line \(\vec{r}=(2 \vec{i}+\vec{j}-3 \vec{k})+t(2 \vec{i}-\vec{j}-\vec{k})\) and the plane x – 2y + 3z + 7 = 0.
Solution:
The equation of the straight line in cartesian form is
\(\frac{x-2}{2}=\frac{y-1}{-1}=\frac{z+3}{-1}\) = λ (say)
Any point on the line is (2λ, + 2, -λ + 1, λ – 3)
The point lies on the plane x – 2y + 3z + 7 = 0
⇒ 2λ, + 2 – 2(-λ + 1) + 3(-λ, – 3) + 7 = 0 .
2λ + 2 + 2λ – 2 – 3λ – 9 + 7 = 0
λ – 2 = 0 ⇒ λ = 2
So 2λ + 2 = 6; -λ + 1 = -1; -λ – 3 = – 5
∴ The required point is (6, -1, -5)

Question 5.
Show that the following planes are at right angles:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 11
Solution:
The normal vectors are
\(\bar{n}_{1}=2 \bar{i}-\bar{j}+\bar{k}\) and \(\vec{n}_{2}=+\vec{i}-\vec{j}-3 \vec{k}\)
\(\vec{n}_{1} \cdot \vec{n}_{2}\) = (2)(1) + (-1)(-1) + (1)(-3) = 2 + 1 – 3 = 0
⇒ \(\vec{n}_{1}\) ⊥r to \(\vec{n}_{2}\), i.e., the normals to the planes are at right angles. So, the planes are at right angles.

Question 6.
The planes \(\vec{r} \cdot(2 \vec{i}+\lambda \vec{j}-3 \vec{k})\) = 10 and \(\vec{r} \cdot(\lambda \vec{i}+3 \vec{j}+\vec{k})\) = 5 are perpendicular. Find λ.
Solution:
Since the planes are perpendicular, the angle between the normals = 90°.
The normals are \(\vec{n}_{1}=2 \vec{i}+\lambda \vec{j}-3 \vec{k}\) and \(\vec{n}_{2}=\lambda \vec{i}+3 \vec{j}+\vec{k}\)
\(\vec{n}_{1} \cdot \vec{n}_{2}\) = 0 [∵ θ = π/2]
⇒ (2)(λ) + (λ)(3) + (-3)(1) = 0 ⇒ 2λ + 3λ – 3 = 0
5λ – 3 = 0 ⇒ 5λ = 3 ⇒ λ = 3/5

Question 7.
Find the angle between the line \(\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{-2}\) and the plane 3x + 4y + z = 0.
Solution:
The angle between the line \(\vec{r}=\vec{a}+t \vec{b}\) and the plane \(\vec{r} \cdot \vec{n}\) = p is given by the formula
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 12

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9

Question 8.
Find the angle between the line \(\vec{r}=\vec{i}+\vec{j}+3 \vec{k}+\lambda(2 \vec{i}+\vec{j}-\vec{k})\) and the plane \(\vec{r} \cdot(\vec{i}+\vec{j})\) = 1,
Solution:
Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.9 13

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