Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 8 Vector Algebra – I Ex 8.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.2

Question 1.

Verify whether the following ratios are direction cosines of some vector or not.

(i) \(\frac{1}{5}, \frac{3}{5}, \frac{4}{5}\)

(ii) \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\)

(iii) \(\frac{4}{3}, 0, \frac{3}{4}\)

Answer:

(i) \(\frac{1}{5}, \frac{3}{5}, \frac{4}{5}\)

[If l, m, n are direction cosines of a vector then l^{2} + m^{2} + n^{2} = 1]

∴ The given ratio \(\frac{1}{5}, \quad \frac{3}{5}, \quad \frac{4}{5}\) do not form the direction cosines of a vector.

(ii) \(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\)

[If l, m, n are direction cosines of a vector then l^{2} + m^{2} + n^{2} = 1]

∴ The given ratio form the direction cosines of a vector.

(iii) \(\frac{4}{3}, 0, \frac{3}{4}\)

[If l, m, n are direction cosines of a vector then l^{2} + m^{2} + n^{2} = 1]

∴ The given ratio do not form the direction cosines of a vector.

Question 2.

Find the direction cosines of a vector whose direction ratios are

(i) 1, 2, 3

(ii) 3, -1 , 3

(iii) 0, 0, 7

Answer:

(i) 1, 2, 3

The given direction ratios are a = 1, b = 2 , c = 3

(ii) 3, – 1, 3

The given direction ratios are a = 3, b = – 1 , c = 3

(iii) 0, 0, 7

The given direction ratios are a = 0, b = 0, c = 7

Question 3.

Find the direction cosines and direction ratios for the following vectors.

(i) 3î – 4ĵ + 8k̂

Answer:

(ii) 3î + ĵ + k̂

Answer:

(iii) ĵ

Answer:

ĵ = 0î + ĵ + 0k̂

The direction ratios of the vector ĵ are (0, 1, 0)

The direction cosines of the vector ĵ are

(0, 1, 0)

Direction ratios = (0, 1, 0)

Direction cosines = (0, 1, 0)

(iv) 5î – 3ĵ – 48k̂

Answer:

(v) 3î – 3k̂ + 4ĵ

Answer:

(vi) î – k̂

Answer:

Question 4.

A triangle is formed by joining the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). Find the direction cosines of the medians.

Answer:

Let ABC be the triangle and D, E, F is the midpoints of the sides BC, CA, AB respectively. Then AD, BE, CF are the medians of ∆ ABC.

Given that the vertices of the triangle are A (1, 0, 0) , B (0, 1, 0 ) and C (0, 0, 1).

Question 5.

If \(\frac{1}{2}, \frac{1}{\sqrt{2}}\), a are the direction cosines of some vector, then find a.

Answer:

Given \(\frac{1}{2}, \frac{1}{\sqrt{2}}\), a are the direction cosines of some vector, then

[If l, m, n are direction cosines of a vector then l^{2} + m^{2} + n^{2} = 1]

Question 6.

If (a, a + b , a + b + c)is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0), then find a set of values of a, b, c.

Answer:

Let A be the point (1, 0, 0) and B be the point (0, 1, 0) (i.e.,) \(\overrightarrow{\mathrm{OA}}=\hat{i}\) and \(\overrightarrow{\mathrm{OB}}=\hat{j}\)

Then \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\hat{j}-\hat{i}=-\hat{i}+\hat{j}\)

= (-1, 1, 0)

= (a, a + b, a + b + c)

⇒ a = -1, a + b = 1 and a + b + c = 0

Now a = -1 ⇒ -1 + b = 1 ;a + b + c = 0

⇒ b = 2; -1 + 2 + c = 0 ⇒ c + 1 = 0

⇒ c = -1

∴ a = -1; b = 2; c = -1.

Note: If we taken \(\overrightarrow{\mathrm{BA}}\) then we get a = 1, b = -2 and c = 1.

Question 7.

Show that the vectors 2î – ĵ + k̂ , 3î – 4ĵ – 4k̂, î – 3ĵ – 5k̂ form a right angled triangle.

Answer:

Question 8.

Find the value of λ for which the vector \(\vec{a}\) = 3î + 2ĵ + 9k̂ and \(\vec{b}\) = î + λĵ + 3k̂ are parallel.

Answer:

Question 9.

Show that the following vectors are coplanar

(i) î – 2ĵ + 3k̂, – 2î + 3ĵ – 4k̂, – ĵ + 2k̂

Answer:

(ii) 5î + 6ĵ + 7k̂, 7î – 8ĵ + 9k̂, 3î + 20ĵ + 5k̂

Answer:

Question 10.

Show that the points whose position vectors 4î + 5ĵ + k̂, – ĵ – k̂, 3î + 9ĵ + 4k̂ and -4î + 4ĵ + 4k̂ are coplanar.

Answer:

Let the given position vectors of the points A, B, C, D be

Equating the like terms on both sides

-4 = 3s – 7t ………… (1)

-6 = 10s – 5t ……….. (2)

-2 = 5s …………. (3)

(3) ⇒ s = \(-\frac{2}{5}\)

Substituting in equation (2) , we have

-6 = 10 × \(-\frac{2}{5}\) – 5t

-6 = -4 – 5t

-6 + 4 = -5t

⇒ -5t = -2

⇒ t = \(\frac{2}{5}\)

Substituting for s and t in equation (1), we have

Question 11.

If \(\vec{a}\) = 2î + 3ĵ – 4k̂ , \(\vec{b}\) = 3î – 4ĵ – 5k̂ and \(\vec{c}\) = – 3î + 2ĵ + 3k̂ , find the magnitude and direction cosines of

(i) \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)

(ii) 3\(\vec{a}\) – 2\(\vec{b}\) + 5\(\vec{c}\)

Answer:

The given vectors are \(\vec{a}\) = 2î + 3ĵ – 4k̂ , \(\vec{b}\) = 3î – 4ĵ – 5k̂ \(\vec{c}\) = – 3î + 2ĵ + 3k̂

(i) \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\)

(ii) 3\(\vec{a}\) – 2\(\vec{b}\) + 5\(\vec{c}\)

Question 12.

The position vectors of the vertices of a triangle are î + 2ĵ + 3k̂ ; 3î – 4ĵ + 5k̂ and – 2î + 3ĵ – 7k̂. Find the perimeter of the triangle.

Answer:

Let A, B, C be the vertices of a triangle ABC.

Given the position vectors of the vertices A, B, C are given by

∴ The required perimeter of the triangle ABC is given by AB + BC + AC = \(\sqrt{44}+\sqrt{218}+\sqrt{110}\)

Question 13.

Find the unit vector parallel to

Answer:

Question 14.

The position vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) of three points satisfy the relation 2 \(\vec{a}\) + 7\(\vec{b}\) + 5\(\vec{c}\) = \(\vec{0}\). Are these points collinear?

Answer:

Question 15.

The position vector of the points P, Q, R, S are î + ĵ + k̂, 2î + 5ĵ, 3î + 2ĵ – 3k̂ and î – 6ĵ – k̂ respectively. Prove that the line and PQ and RS are parallel.

Answer:

Given that the position vector of the given points P, Q, R, S are

Question 16.

Find the value or values of m for which m (î + ĵ + k̂) is a unit vector.

Answer:

The given vector is m (î + ĵ + k̂)

Given that it is a unit vector.

∴ |m (i + j + k)| = 1

|m (i + j + k)| = 1

m (± \(\sqrt{1^{2}+1^{2}+1^{2}}\)) = 1

m = ± \(\frac{1}{\sqrt{3}}\)

Question 17.

Show that points A (1, 1, 1), B (1, 2, 3), and C (2, -1, 1 ) are vertices of an isosceles triangle.

Answer:

The given vertices of a triangle are

A(1, 1, 1) ,B(1 , 2, 3) and C (2, – 1, 1)

Position vector of A is \(\overrightarrow{\mathrm{OA}}\) = î + ĵ + k̂

Position vector of B is \(\overrightarrow{\mathrm{OB}}\) = î + 2ĵ + 3k̂

Position vector of C is \(\overrightarrow{\mathrm{OC}}\) = 2î – ĵ + k̂

∴ A, B, C are not collinear.

Hence A, B, C form a triangle.

AB = CA = √5

∴ ∆ABC is an isosceles triangle.