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You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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)

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

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

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

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

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

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:

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 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

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

∴ The required point is (2, 0, 0)

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:

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

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:

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.8

Question 1.
Show that the straight lines and are coplanar. Find the vector equation of the plane in which they lie.
Solution:

Question 2.
Show that lines are coplanar. Also, find the plane containing these lines.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (2, 3, 4) and (x₂, y₂, z₂) = (1, 4, 5)
(b₁, b₂, b₃) = (1, 1, 3) and (d₁, d₂, d₃) = (-3, 2, 1)
Condition for coplanarity

(x – 2)[1 – 6] – (y – 3)[1 + 9] + (z – 4)[2 + 3] = 0
-5(x – 2) – 10(y – 3) + 5(z – 4) = 0
-5x + 10 – 10y + 30 + 5z – 20 = 0
-5x – 10y + 5z + 20 = 0
(÷ by -5) ⇒ x + 2y – 2z – 4 = 0

Question 3.
If the straight lines are coplanar, find the district real values of m.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (1, 2, 3) and (x₂, y₂, z₂) = (3, 2, 1)
(b₁, b₂, b₃) = (1, 2, m²) and (d₁, d₂, d₃) = (1, m², 2)
Condition for coplanarity
\(\left|\begin{array}{ccc}{x_{2}-x_{1}} & {y_{2}-y_{1}} & {z_{2}-z_{1}} \\ {b_{1}} & {b_{2}} & {b_{3}} \\ {d_{1}} & {d_{2}} & {d_{3}}\end{array}\right|=0\)

Question 4.
If the straight lines are coplanar, find λ and equation of the planes containing these two lines.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (1, -1, 0) and (x₂, y₂, z₂) = (-1, -1, 0)
(b₁, b₂, b₃) = (2, λ, 2) and (d₁, d₂, d₃) = (5, 2, λ)
Condition for coplanarity

\(\left|\begin{array}{ccc}{x-1} & {y+1} & {z} \\ {2} & {-2} & {2} \\ {5} & {2} & {-2}\end{array}\right|\) = 0
(x – 1)[0] – (y + 1)[-4 – 10] + z[4 + 10] = 0
14(y + 1) + 14z = 0 ⇒ 14y + 14 + 14z = 0
(÷ by 14) ⇒ y + z + 1 = 0

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.8 Additional Problems

Question 1.
Show that the straight lines.

are coplanar. Find the vector equation of the plane in which they lie.
Solution:

Question 2.
If the straight lines are coplanar. Find λ.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (1, 1, 1) and (b₁, b₂, b₃) = (1, λ, 1)
(x₂, y₂, z₂) = (0, 4, 2) and (d₁, d₂, d₃) = (2, λ, 3)
Condition for coplanarity
\(\left|\begin{array}{ccc}{x_{2}-x_{1}} & {y_{2}-y_{1}} & {z_{2}-z_{1}} \\ {b_{1}} & {b_{2}} & {b_{3}} \\ {d_{1}} & {d_{2}} & {d_{3}}\end{array}\right|\) = 0
\(\left|\begin{array}{ccc}{-1} & {3} & {1} \\ {1} & {\lambda} & {1} \\ {2} & {\lambda} & {3}\end{array}\right|\) = 0
-1(3λ – λ) – 3(3 – 2) + 1(λ – 2λ) = 0 ⇒ -2λ – 3 – λ = 0
-3λ = 3 ⇒ λ = -1

Question 3.
If the lines are coplanar, then find the value of k.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (2, 3, 4) and (b₁, b₂, b₃) = (1, 1, -1)
(x₂, y₂, z₂) = (1, 4, 5) and (d₁, d₂, d₃) = (k, 2, 1)
Condition for coplanarity

-1(1 + 2k) – 1(1 + k²) + 1(2 – k) = 0
-1 – 2k – 1 – k² + 2 – k = 0
-k² – 3k = 0
k² + 3k = 0
k(k + 3) = 0
k = 0 or k = -3

Question 4.
Show that the lines are coplanar, Also find the equation of the plane containing these two lines.
Solution:
From the lines we have,
(x₁, y₁, z₁) = (-3, 1, 5) and (b₁, b₂, b₃) = (-3, 1, 5)
(x₂, y₂, z₂) = (-1, 2, 5) and (d₁, d₂, d₃) = (-1, 2, 5)
Condition for coplanarity

Given two lines are coplanar

(x + 3)[5 – 10] – (y – 1)[-15 + 5] + (z – 5)[-6 + 1] = 0
5(x + 3) + 10(y – 1) – 5(z – 5) = 0
(÷ by 5) ⇒ (x + 3) -2(y – 1) + (z – 5) = 0
x + 3 – 2y + 2 + z – 5 = 0
x – 2y + z = 0
or

(x + 1)(5 – 10) – (y – 2)(-15 + 5) + (z – 5)(-6 + 1) = 0
-5(x + 1) + 10(y – 2) – 5(z – 5) = 0
(x + 1) – 2(y – 2) + (z – 5) = 0
x + 1 – 2y + 4 + z – 5 = 0
x – 2y + z = 0

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.7

Question 1.
Find the non-parametric form of vector equation, and Cartesian equation of the plane passing through the point (2, 3, 6) and parallel to the straight lines \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-3}{1}\) and \(\frac{x+3}{2}=\frac{y-3}{-5}=\frac{z+1}{-3}\)
Solution:

Question 2.
Find the parametric form of vector equation, and Cartesian equations of the plane passing through the points (2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 9.
Solution:
The required plane passes through the points

Question 3.
Find the parametric form of vector equation and Cartesian equations of the plane passing through the points (2, 2, 1), (1, -2, 3) and parallel to the straight line passing through the points (2, 1, -3) and (-1, 5, -8).
Solution:
Equation of the straight line passing through the points (2, 1, -3) and (-1, 5, -8) is

(x – 2) [20 – 8] – (y – 2) [5 + 6] + (z – 1) [-4 – 12] = 0
12(x – 2) – 11(y – 2) – 16(z – 1) = 0
12x – 24 – 11y + 22 – 16z + 16 = 0
12x – 11y – 16z + 14 = 0

Question 4.
Find the non-parametric form of vector equation of the plane passing through the point (1, -2, 4) and perpendicular to the plane x + 2y – 3z = 11 and parallel to the line \(\frac{x+7}{3}=\frac{y+3}{-1}=\frac{z}{1}\)
Solution:
The required plane passing through the point \(\vec{a}=\vec{i}-2 \vec{j}+4 \vec{k}\) and parallel to the plane \(\vec{b}=\vec{i}+2 \vec{j}-3 \vec{k}\) and parallel to the line \(\vec{c}=3 \vec{i}-\vec{j}+\vec{k}\)

Question 5.
Find the parametric form of vector equation, and Cartesian equations of the plane containing the line \(\vec{r}=(\hat{i}-\hat{j}+3 \hat{k})+t(2 \hat{i}-\hat{j}+4 \hat{k})\) and perpendicular to the plane \(\vec{r} \cdot(\hat{i}+2 \hat{j}+\hat{k})\) = 8.
Solution:
The required plane passing through the point \(\vec{a}=\vec{i}-\vec{j}+3 \vec{k}\) and parallel to \(\vec{b}=2 \vec{i}-\vec{j}+4 \vec{k}\) and \(\vec{c}=\vec{i}+2 \vec{j}+\vec{k}\)
Parametric form of vector equation

(x – 1) [-1 – 8] – (y + 1)[z – 4] + (z – 3) [4 + 1] = 0
-9(x- 1) + 2(y + 1) + 5(z- 3) = 0
-9x + 9 + 2y + 2 + 5z – 15 = 0
-9x + 2y + 5z – 4 = 0
9x – 2y – 5z + 4 = 0

Question 6.
Find the parametric vector, non-parametric vector and Cartesian form of the equation of the plane passing through the point (3, 6, -2), (-1, -2, 6) and (6, 4, -2).
Solution:
The required plane passing through the points

Question 7.
Find the non-parametric form of vector equation, and Cartesian equations of the plane

Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.7 Additional Problem

Question 1.
Find the vector and cartesian equations of the plane containing the line \(\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{3}\) and parallel to the line \(\frac{x+1}{3}=\frac{y-1}{2}=\frac{z+1}{1}\)
Solution:
The required plane passes through the point A(2, 2, 1) and parallel to

i.e., (x – 2)[3 – 6] – (y – 2)[2 – 9] + (z – 1)[4 – 9] = 0
i.e., (x – 2)(-3) + (y – 2)(7) – 5(z – 1) = 0
-3x + 6 + 7y – 14 – 5z + 5 = 0
-3x + 7y – 5z – 3 = 0 i.e. 3x – 7y + 5z + 3 = 0

Question 2.
Find the vector and cartesian equation of the plane passing through the point (1, 3, 2) and parallel to the lines
Solution:
The required plane passes through the point A = (1, 3, 2) and parallel to the vectors

i.e. (x – 1)(-2 – 6) – (y – 3)(4 – 3) + (z – 2)(4 + 1) = 0
—8(x – 1) – 1(y – 3) + 5(z – 2) = 0
-8x + 8 – y + 3 + 5z – 10 = 0
-8x – y + 5z + 1 = 0 i.e. 8x + y – 5z – 1 = 0

Question 3.
Find the vector and cartesian equations of the plane passing through the point (-1, 3, 2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + y + 2z = 8
Solution:
The normal vector to the given planes x + 2y + 2z = 5 and 3x + y + 2z = 8 are respectively \(\vec{i}+2 \vec{j}+2 \vec{k}\) and \(3 \vec{i}+\vec{j}+2 \vec{k}\). These vectors are parallel to the required plane
The required plane passes through the point A(-1, 3, 2) and parallel to the vectors

i.e., (x + 1)(4 – 2) – (y – 3)(2 – 6) + (z – 2)(1 – 6) = 0
2(x + 1) + 4(y – 3) – 5(z – 2) = 0
2x + 2 + 4y – 12 – 5z + 10 = 0
2x + 4y – 5z = 0

Question 4.
Find the vector and cartesian equations of the plane passing through the points A(1, -2, 3) and B(-1, 2, -1) and is parallel to the line \(\frac{x-2}{2}=\frac{y+1}{3}=\frac{z-1}{4}\)
Solution:
The vector equation of the plane is

i.e, (x – 1)[16 + 12] – (y + 2)(-8 + 8) + (z – 3)(-6 – 8) = 0
28(x – 1) – 14(z – 3) = 0
28x – 28 – 14z + 42 = 0
28x – 14z + 14 = 0
(÷ by 14) ⇒ 2x – z + 1 = 0

Question 5.
Find the vector and cartesian equations of the plane through the points (1, 2, 3) and (2,3,1) and perpendicular to the plane 3x – 2y + 4z – 5 = 0.
Solution:
The vector normal to the plane 3x – 2y + 4z – 5 = 0 is \(3 \vec{i}-2 \vec{j}+4 \vec{k}\)
The required plane is parallel to the vector \(3 \vec{i}-2 \vec{j}+4 \vec{k}\)
∴ Vector equation of the plane passing through two points and parallel to one vector is

i.e. (x – 1)[4 – 4] – (y – 2)[4 + 6] + (z – 3)[-2 – 3] = 0
-10(y – 2) – 5(z – 3) = 0
-10y + 20 – 5z + 15 = 0
-10y – 5z + 35 = 0
10y + 5z – 35 = 0
(÷ by 5) ⇒ 2y + z – 1 = 0

Question 6.
Find the vector and cartesian equations of the plane containing the line \(\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{-2}\) and passing through the point (-1, 1, -1).
Solution:
The required plane passes through the points A(-1, 1,-1) and B(2, 2, 1) and parallel to the vectors \(\vec{c}=2 \vec{i}+3 \vec{j}-2 \vec{k}\) .
The required equation is

i.e. (x + 1)[-2 – 6] – (y – 1)[-6 – 4] + (z + 1)[9 – 2] = 0
-8(x + 1) + 10(y – 1) + 7(z + 1) = 0
-8x – 8 + 10y – 10 + 7z + 7 = 0
-8x + 10y + 7z – 11 = 0
i.e., 8x – 10y – 7z + 11 = 0

Question 7.
Find the vector and cartesian equations of the plane passing through the points with position vectors .
Solution:
Vector equation of the plane passing through three given non-collinear points is

i.e. (x – 3)[6 – 12] – (y – 4)[1 + 12] + (z – 2)[4 + 24] = 0
-6(x – 3) – 13(y – 4) + 28(z – 2) = 0
-6x + 18 – 13y + 52 + 28z – 56 = 0
-6x – 13y + 28z + 14 = 0 .
i.e. 6x + 13y – 28z – 14 = 0

Question 8.
Derive the equation of the plane in the intercept form.
Solution:
Let the required plane makes intercepts on X, Y, Z-axes respectively as a, b and c.
i.e. A = (a, 0, 0)
B = (0, b, 0)
C = (0, 0, c)
The equation of the plane passing through three non-collinear points A, B and C is

Question 9.
Find the cartesian form of the following plane:

Solution:

x(0 + 2) – (y – 3) (1 + 4) + z(-1) = 0
2x – 5(y – 3) – z = 0
2x – 5y + 15 – z = 0
2x – 5y – z + 15 = 0
which is the cartesian equation of the plane.

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.


Solution:
(a) \(\frac{\pi}{6}\)
Hint:

Question 2.


Solution:
(c) \(\frac{5}{2}\)
Hint:

Question 3.


Solution:
(c) 0
Hint:

Question 4.


Solution:
(d) \(\frac{2}{3}\)
Hint:
It is an even function

Question 5.

(b) 2π
(c) 3π
(d) 4π
Solution:
(d) 4π
Hint:

Question 6.

(a) 4
(b) 3
(c) 2
(d) 0
Solution:
(c) 2
Hint:

Question 7.

(a) cos x – x sin x
(b) sin x + x cos x
(c) x cos x
(d) x sin x
Solution:
(c) x cos x
Hint:

Question 8.
The area between y² = 4x and its latus rectum is ………

Solution:
(c) \(\frac{8}{3}\)
Hint:

Question 9.


Solution:
(b) \(\frac{1}{10100}\)

Question 10.


Solution:
(a) \(\frac{\pi}{2}\)

Question 11.

(a) 10
(b) 5
(c) 8
(d) 9
Solution:
(d) 9
Hint:

Question 12.


Solution:
(b) \(\frac{2}{9}\)
Hint:

Question 13.


Solution:
\(\frac{3 \pi}{8}\)
Hint:

Question 14.


Solution:
(d) \(\frac{2}{27}\)
Hint:

Question 15.

(a) 4
(b) 1
(c) 3
(d) 2
Solution:
(d) 2
Hint:

Question 16.
The volume of solid of revolution of the region bounded by y² = x(a – x) about x-axis is ……..

Solution:
(d) \(\frac{\pi a^{3}}{6}\)
Hint:

Question 17.

(a) 3
(b) 6
(c) 9
(d) 5
Solution:
(c) 9
Hint:

Question 18.


Solution:
(d) \(\frac{\pi^{2}}{4}-2\)
Hint:

Question 19.


Solution:
(b) \(\frac{3 \pi a^{4}}{16}\)
Hint:

Question 20.


Solution:
(a) \(\frac{1}{2}\)
Hint:

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.10 Additional Problems

Choose the correct or the most suitable answer from the given four alternatives:

Question 1.
The area bounded by the line y = x, the x – axis, the ordinates x = 1,x = 2 is …….

Solution:
(a) \(\frac{3}{2}\)
Hint:

Question 2.
The area of the region bounded by the graph of y = sin x and y = cos x between x = 0 and x = \(\frac{\pi}{4}\) is ……..

Solution:
(b) \(\sqrt{2}-1\)
Hint:

Question 3.
The area between the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 and its auxiliary circle is …….
(a) πb(a – b)
(b) 2πa(a – b)
(c) πa(a – b)
(d) 2πb(a – b)
Solution:
(c) πa(a – b)
Hint:

Question 4.
The area bounded by the parabola y² = x and its latus rectum is ……..

Solution:
(b) \(\frac{1}{6}\)
Hint:

Question 5.
The volume of the solid obtained by revolving \(\frac{x^{2}}{9}+\frac{y^{2}}{16}\) = 1 about the minor axis is …….
(a) 48π
(b) 64π
(c) 32π
(d) 128π
Solution:
(b) 64π
Hint:

Question 6.
The volume, when the curve y = \(\sqrt{3+x^{2}}\) from x = 0 to x = 4 is rotated about x – axis is ……

Solution:
(c) \(\frac{100}{3} \pi\)
Hint:

Question 7.
The volume generated when the region bounded by y = x, y = 1, x = 0 is rotated about y – axis is ……….

Solution:
(c) \(\frac{\pi}{3}\)
Hint:

Question 8.
Volume of solid obtained by revolving the area of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 about major and minor axes are in the ratio …….
(a) b² : a²
(b) a² : b²
(c) a : b
(d) b : a
Solution:
(d) b : a
Hint:

Question 9.
The volume generated by rotating the triangle with vertices at (0, 0), (3, 0) and (3, 3) about x-axis is …….
(a) 18π
(b) 2π
(c) 36π
(d) 9π
Solution:
(d) 9π
Hint:

Question 10.
The length of the arc of the curve is …….
(a) 48
(b) 24
(c) 12
(d) 96
Solution:
(a) 48
Hint:
Length of the arc of the curve = 6a

∴ Required length = 6a = 6 × 8 = 48 units.

Question 11.
The surface area of the solid of revolution of the region bounded by y = 2x, x = 0 and x = 2 about x-axis is ……

Solution:
(a) \(8 \sqrt{5} \pi\)
Hint:

Question 12.
The curved surface area of a sphere of radius 5, intercepted between two parallel planes of distance 2 and 4 from the centre is ……
(a) 20π
(b) 40π
(c) 10π
(d) 30π
Solution:
(a) 20π
Hint:
The curved surface area of a sphere of radius r intercepted between two parallel planes at a distance a and b from the centre of the sphere is 2πr (b – a)
Given radius, r = 5; a = 2; b = 4
Required surface area = 2πr (b – a)
= 2π × 5 × (4 – 2) = 20π sq. units

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.6

Question 1.
Find a parametric form of vector equation of a plane which is at a distance of 7 units from the origin having 3, -4, 5 as direction ratios of a normal to it.
Solution:
Given p = 7

Question 2.
Find the direction cosines of the normal to the plane 12x + 3y – 4z = 65. Also, find the non-parametric form of vector equation of a plane and the length of the perpendicular to the plane from the origin.
Solution:
12x + 3y – 4z = 65

(iii) Length of the perpendicular to the plane from the origin is 5 units.

Question 3.
Find the vector and Cartesian equation of the plane passing through the point with position vector \(2 \hat{i}+6 \hat{j}+3 \hat{k}\) and normal to the vector \(\hat{i}+3 \hat{j}+5 \hat{k}\)
Solution:

Question 4.
A plane passes through the point (-1, 1, 2) and the normal to the plane of magnitude \(3 \sqrt{3}\) makes equal acute angles with the coordinate axes. Find the equation of the plane.
Solution:
Given magnitude = \(3 \sqrt{3}\) and \(\vec{a}=-\vec{i}+\vec{j}+2 \vec{k}\)
Then, the normal vector makes equal acute angle with the coordinate axes.
We know that cos² α + cos² β + cos² γ = 1 (But α = β = γ)

Question 5.
Find the intercepts cut off by the plane \(\vec{r} \cdot(6 \hat{i}+4 \hat{j}-3 \hat{k})\) = 12 on the coordinate axes.
Solution:

x-intercept = 2; y-intercept = 3; z-intercept = -4

Question 6.
If a plane meets the coordinate axes at A, B, C such that the centroid of the triangle ABC is the point (u, v, w), find the equation of the plane.
Solution:
Let A (a, 0, 0), B(0, b, 0), C(0, 0, c)
centroid of ∆ABC = \(\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\)

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.6 Additional Problems

Question 1.
Find the vector and cartesian equations of a plane which is at a distance of 18 units from the origin and which is normal to the vector \(2 \vec{i}+7 \vec{j}+8 \vec{k}\)
Solution:

Question 2.
Find the unit vector to the plane 2x – y + 2z = 5.
Solution:
Writing the plane in normal form we get,

Question 3.
Find the length of the perpendicular from the origin to the plane \(\vec{r} \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})\) = 26.
Solution:
Taking the equation of the plane in cartesian form we get,
\((x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})\) = 26
i.e., 3x + 4y+ 12z – 26 = 0
The length of the perpendicular from (0, 0, 0) to the above plane is
\(\pm \frac{-26}{\sqrt{9+16+144}}=\frac{+26}{13}\) = 2 units

Question 4.
The foot of the perpendicular drawn from the origin to a plane is (8, -4, 3). Find the equation of the plane.
Solution:
The required plane passes through the point A(8, -4, 3) and is perpendicular to \(\overrightarrow{\mathrm{OA}}\) .

The cartesian equation is 8x – 4y + 3z = 89.

Question 5.
Find the equation of the plane through the point whose position vector is \(2 \vec{i}-\vec{j}+\vec{k}\) and perpendicular to the vector \(4 \vec{i}+2 \vec{j}-3 \vec{k}\).
Solution:
The required plane is perpendicular to \(4 \vec{i}+2 \vec{j}-3 \vec{k}\)
So, it is parallel to the plane 4x + 2y – 3z = k
∴ the equation of the plane is 4x + 2y – 3z = k
The plane passes through the point (2, -1, 1)
⇒ (4)(2) + 2(-1) – 3(1) = λ i.e. λ = 8 – 2 – 3 = 3
So, the equation of the plane is 4x + 2y – 3z = 3.

Question 6.
Find the vector and cartesian equations of the plane passing through the point (2, -1, 4) and parallel to the plane \(\vec{r} \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})\) = 7.
Solution:
The given plane is \(\vec{r} \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})\) = 7
i e. \((x \vec{i}+y \vec{j}+z \vec{k}) \cdot(4 \vec{i}-12 \vec{j}-3 \vec{k})\) = 7
i.e. 4x – 12y – 3z = 1
The required plane is parallel to the above plane. So, the equation of the required plane is 4x – 12y – 3z – k. The plane passes through (2, -1, 4).
⇒ 4(2) – 12(-1) – 3(4) = k i.e. k = 8 + 12 – 12 = 8
So, the equation of the plane is 4x – 12y – 3z = 8.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.5

Question 1.
Find the parametric form of vector equation and Cartesian equations of a straight line passing through (5, 2, 8) and is perpendicular to the straight lines

Solution:

∴ This’ vector is perpendicular to both the given straight lines.
∴ The required straight line is
\(\vec{r}=\vec{a}+t(\vec{b} \times \vec{d})\)

Question 2.
Show that the lines are skew lines and hence find the shortest distance between them.
Solution:

Question 3.
If the two lines intersect at a point, find the value of m.
Solution:

Question 4.
Show that the lines \(\frac{x-3}{3}=\frac{y-3}{-1}\), z – 1 = 0 and \(\frac{x-6}{2}=\frac{z-1}{3}\), y – 2 = 0 intersect. Also find the point of intersection
Solution:

Any point on the Second line

∴ The required point of intersection is (6, 2, 1)

Question 5.
Show that the straight lines x + 1 = 2y = -12z and x = y + 2 = 6z – 6 are skew and hence find the shortest distance between them.
Solution:

Question 6.
Find the parametric form of vector equation of the straight line passing through (-1, 2, 1) and parallel to the straight line \(\vec{r}=(2 \hat{i}+3 \hat{j}-\hat{k})+t(\hat{i}-2 \hat{j}+\hat{k})\) and hence find the shortest distance between the lines.
Solution:

Question 7.
Find the foot of the perpendicular drawn from the point (5, 4, 2) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}\). Also, find the equation of the perpendicular.
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.5 Additional Problems

Question 1.
Find the shortest distance between the parallel line

Solution:

Question 2.
Show that the two lines \(\vec{r}=(\vec{i}-\vec{j})+t(2 \vec{i}+\vec{k})\) and \(\vec{r}=(2 \vec{i}-\vec{j})+s(\vec{i}+\vec{j}-\vec{k})\) skew lines and find the distance between them.
Solution:

Question 3.
Show that the lines intersect and hence find the point of intersection
Solution:

Any point on this line is of the form (2µ + 4, 0, 3µ – 1)
Since they are intersecting, for some λ, µ
(3λ + 1, – λ + 1,- 1) = (2µ + 4, 0, 3µ – 1) ⇒ λ = 1 and µ = 0
To find the point of intersection either take λ = 1 or µ = 0
∴ The point of intersection is (4, 0, – 1),

Question 4.
Find the shortest distance between the skew lines.

Solution:
Compare the given equation with \(\vec{r}=\vec{a}_{1}+t \vec{u}\) and \(\vec{r}=\vec{a}_{2}+s \vec{v}\)

Question 5.
Find the shortest distance between the parallel lines.

Solution:

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9

Question 1.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = 2x², y = 0 and x = 1.
Solution:

Question 2.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = e^-2x y = 0, x = 0 and x = 1.
Solution:

Question 3.
Find, by integration, the volume of the solid generated by revolving about the y-axis, the region enclosed by x² = 1 + y and y = 3.
Solution:

Question 4.
The region enclosed between the graphs of y = x and y = x² is denoted by R, Find the volume generated when R is rotated through 360° about x – axis.
Solution:
To find points of intersection, solving y = x² and y = x, we get (0, 0) and (1, 1)

Question 5.
Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown in the Figure.
Solution:

Question 6.
A watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major-axis 20 cm and minor-axis 10 cm about its major-axis. Find its volume using integration.
Solution:
From the given data a = 10 cm and b = 5 cm
Equation of the Ellipse

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.9 Additional Questions

Question 1.
Find the volume of the solid that results when the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (a > b > 0) is revolved about the minor axis.
Solution:
Volume of the solid is obtained by revolving the right side of the curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) about the y-axis.
Limits for y is obtained by putting x = 0 ⇒ y² = b² ⇒ y = ±b.

Question 2.
Find the volume of the solid generated when the region enclosed by y = \(\sqrt{x}\), y = 2 and x = 0 is revolved about the y – axis.
Solution:
Since the solid is generated by revolving about the y-axis, rewrite y = \(\sqrt{x}\) as x = y².
Taking the limits for y, y = 0 and y = 2 (Putting x = 0 in x = y², we get y = 0)

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4

Question 1.
Find the non-parametric form of vector equation and Cartesian equations of the straight line passing through the point with position vector \(4 \hat{i}+3 \hat{j}-7 \hat{k}\) and parallel to the vector \(2 \hat{i}-6 \hat{j}+7 \hat{k}\)
Solution:

Question 2.
Find the parametric form of vector equation and Cartesian equations of the straight line passing through the point (-2, 3, 4) and parallel to the straight line \(\frac{x-1}{-4}=\frac{y+3}{5}=\frac{8-z}{6}\)
Solution:

Question 3.
Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
Solution:
Given straight line passing through the points (6, 7, 4) and (8, 4, 9).
Direction ratio of the straight line joining these two points 2, -3, -5.
Cartesian equation:

(ii) The straight Line cuts yz-plane
So we get x = 0
2t + 6 = 0 ⇒ 2t = -6
t = -3
-3t + 7 = -3 (-3) + 7 = 9 + 7 = 16
5t + 4 = 5(-3) + 4 = -15 + 4 = -11
The required point (0, 16, -11).

Question 4.
Find the direction cosines of the straight line passing through the points (5, 6, 7) and (7,9,13). Also, find the parametric form of vector equation and Cartesian equations of the straight line passing through two given points.
Solution:
Given straight line passing through the points (5, 6, 7) and (7, 9, 13)
∴ d.r.s : 2, 3, 6

Note: Selection of \(\vec{a}\) and \(\vec{b}\) is your choice.

Question 5.
Find the acute angle between the following lines.

(iii) 2x = 3y = -z and 6x = -y = -4z
Solution:

Question 6.
The vertices of ∆ABC are A(7, 2, 1), B(6, 0, 3), and C(4, 2, 4) . Find ∠ABC.
Solution:

Question 7.
If the straight line joining the points (2, 1, 4) and (a – 1, 4, -1) is parallel to the line joining the points (0, 2, b – 1) and (5, 3, -2), find the values of a and b.
Solution:

Question 8.
If the straight lines are perpendicular to each other, find the value of m.
Solution:

Question 9.
Show that the points (2, 3, 4),(-1, 4, 5) and (8, 1, 2) are collinear.
Solution:
Given points are (2, 3, 4), (-1, 4, 5) and (8, 1, 2) Equation of the line joining of the first and second point is
\(\frac{x-2}{-3}=\frac{y-3}{1}=\frac{z-4}{1}\) = m (say)
(-3m + 2, m + 3, m + 4)
On putting m = -2, we get the third point is (8, 1, 2)
∴ Given points are collinear.

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.4 Additional Problems

Question 1.
Find the vector and cartesian equations of the straight line passing through the point A with position vector \(3 \vec{i}-\vec{j}+4 \vec{k}\) and parallel to the vector \(-5 \vec{i}+7 \vec{j}+3 \vec{k}\).
Solution:
We know that vector equation of the line through the point with position vector \(\vec{a}\) and parallel to \(\vec{v}\) is given by \(\vec{r}=\vec{a}+t \vec{v}\) where t is a scalar.
Here \(\vec{a}=3 \vec{i}-\vec{j}+4 \vec{k}\) and \(\vec{v}=-5 \vec{i}+7 \vec{j}+3 \vec{k}\)
Vector equation of the line is
\(\vec{r}=(3 \vec{i}-\vec{j}+4 \vec{k})+t(-5 \vec{i}+7 \vec{j}+3 \vec{k})\) ………………. (1)
The cartesian equation of the line passing through (xp yx, zx) and parallel to a vector whose d.r.s are l, m, n is

Question 2.
Find the vector and cartesian equations of the straight line passing through the points (- 5, 2, 3) and (4, – 3, 6).
Solution:
Vector equation of the straight line passing through two points with position vectors \(\vec{a}\) and \(\vec{b}\) is given by

Question 3.
Find the angle between the lines.

Solution:
Let the given lines be in the direction of \(\vec{u}\) and \(\vec{v}\)

Question 4.
Find the angle between the following lines
Solution:
Angle between two lines is the same as angle between their parallel vectors.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8

Question 1.
Find the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x-axis.
Solution:

Question 2.
Find the area of the region bounded by 2x – y + 1 = 0, y = – 1, y = 3 and y – axis.
Solution:
2x – y + 1 = 0

To find further limit put x = 0, we get y = 1

Question 3.
Find the area of the region bounded by the curve 2 + x – x² + y = 0, x – axis, x = – 3 and x = 3.
Solution:

Question 4.
Find the area of the region bounded by the line y = 2x + 5 and the parabola y = x² – 2x.
Solution:
To find point of intersection of the curves
y = 2x + 5 and y = x² – 2x we get (-1, 3) and (5, 15)

Question 5.
Find the area of the region bounded between the curves y = sin x and y = cos x and the lines x = 0 and x = π.
Solution:
To find the points of intersection,

Question 6.
Find the area of the region bounded by y = tan x, y = cot x and the line x = 0, x = \(\frac{\pi}{2}\), 0
Solution:
To find the points of intersection of these two curves between 0 to \(\frac{\pi}{2}\) is \(\frac{\pi}{4}\)

Question 7.
Find the area of the region bounded by parabola y² = x and the line y = x – 2
Solution:
To find the points of intersection solve the two equations y² = x and y = x – 2

Question 8.
Father of a family wishes to divide his square field bounded by x = 0, x = 4 , y = 4 and y = 0 along the curve y² = 4x and x² = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.
Solution:
To find the points of intersection of the two curves, y² = 4x and x² = 4y are (0, 0) and (4, 4).
Area of the square field = 4 × 4 = 16 sq. units

So, the remaining each of the two parts must be \(\frac{16}{3}\) sq.units.
∴ Yes, It is possible to divide the square field into three equal parts.

Question 9.
The curves = (x – 2)² + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.
Solution:
y = (x – 2)² + 1



∴ x = 2 is a minimum point
∴ The point P is (2, 1)
But slope of PQ is 2
∴ Equation of the chord PQ
y – y₁ = m(x – x₁)
y – 1 = 2 (x – 2)
y – 1 = 2x – 4
y = 2x – 3
On solving the curve and line we get the point Q(4, 5)

Question 10.
Find the area of the region common to the circle x² + y² = 16 and the parabola y² = 6x.
Solution:
To find points of intersection of x² + y² = 16 and y² = 6x are (2, \(2 \sqrt{3})\)) and (2, –\(2 \sqrt{3})\))
Due to symmetrical property,

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8 Additional Problems

Question 1.
Find the area of the region enclosed by y² = x and y = x – 2.
Solution:
The points of intersection of the parabola y² = x and the line y = x – 2 are (1, -1) and (4, 2)

To compute the region [shown in figure] by integrating with respect to x, we would have to split the region into two parts, because the equation of the lower boundary changes at x = 1. However if we integrate with respect toy no splitting is necessary.

Question 2.
Find the area bounded by the curve y = x³ and the line y = x.
Solution:
The line y = x lies above the curve y = x³ in the first quadrant and y = x³ lies above the line y = x in the third quadrant. To get the points of intersection, solve the curves y = x³, y = x ⇒ x³ = x. We get x = {0, ± 1}

Question 3.
Find the area of the loop of the curve 3ay² = x (x – a)².
Solution:
Put y = 0; we get x = 0, a
It meets the x – axis at x = 0 and x = a
∴ Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x – axis.

Question 4.
Find the area between the line y = x + 1 and the curve y = x² – 1.
Solution:
To get the points of intersection of the curves we should solve the equations y = x +1 and y = x² – 1.
we get, x² – 1 = x + 1
x² – x – 2 = 0 ⇒ (x – 2)(x + 1) = 0
x = – 1 or x = 2
∴ The line intersects the curve at x = – 1 and x = 2.

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.3

Question 1.

Solution:

Question 2.
For any vector \(\vec{a}\), prove that
Solution:

Question 3.
Prove that \([\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}]\) = 0
Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.
If \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) are coplanar vectors, show that \((\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\overrightarrow{0}\)
Solution:

Question 7.
If , find the values of l, m, n
Solution:

On solving (3) & (4)

Question 8.
If \(\hat{a}, \hat{b}, \hat{c}\) are three unit vectors such that \(\hat{b} \text { and } \hat{c}\) are non-parallel and \(\hat{a} \times(\hat{b} \times \hat{c})=\frac{1}{2} \hat{b}\), find the angle between \(\hat{a}\) and \(\hat{c}\).
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.3 Additional Problems

Question 1.
If and show that they are not equal.
Solution:

Question 2.

Solution: