**Chapter 5 Vectors Ex 5.3**

## Chapter 5 Vectors Ex 5.3

**Question 1.**

**Question 2.**

∴ LHS = RHS

Hence, (a¯ + b¯)

^{2}= (a¯ – b¯)

^{2}.

**Question 3.**

∴ c < 0.

Hence, the angle between a and b is obtuse if c < 0.

**Question 4.**

**Question 5.Suppose that all sides of a quadrilateral are equal in length and opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular.Solution:**

**Question 6.**

= -3 + 2 + 1

= 0

**Question 7.Find the angle P of the triangle whose vertices are P(0, -1, -2), Q(3, 1, 4) and R(5, 7, 1).Solution:**

∴ P = 45°

**Question 8.**

Solution:

**Question 9.Prove by vector method that the angle subtended on semicircle is a right angle.Solution:**

Hence, the angle subtended on a semicircle is the right angle.

**Question 10.If a vector has direction angles 45ºand 60º find the third direction angle.Solution:**

Let α = 45°, β = 60°

We have to find γ.

∴ cos

^{2}α + cos

^{2}β + cos

^{2}γ = 1

∴ cos

^{2}45° + cos

^{2}60° + cos

^{2}γ = 1

**Question 11.If a line makes angles 90º, 135º, 45º with the X, Y and Z axes respectively, then find its direction cosines.Solution:**

Let l, m, n be the direction cosines of the line.

Then l = cos α, m = cos β, n = cos γ

Here, α = 90°, β = 135° and γ = 45°

∴ l = cos 90° = 0

**Question 12.If a line has the direction ratios, 4, -12, 18 then find its direction cosines.Solution:**

The direction ratios of the line are a = 4, b = -12, c = 18.

Let l, m, n be the direction cosines of the line.

**Question 13.**

The coordinates of the points which are at a distance of d units from the point (x_{1}, y_{1}, z_{1}) are given by (x_{1} ± ld, y_{1} ± md, z_{1} ± nd)

i.e. (4 – 4, 1 + 4, 5 + 2) and (4 + 4, 1 – 4, 5 – 2)

i.e. (0, 5, 7) and (8, -3, 3).

**Question 14.Find the angle between the lines whose direction cosines l, m, n satisfy the equations 5l + m + 3n = 0 and 5mn – 2nl + 6lm = 0.Solution:**
Given, 5l + m + 3n = 0 …(1)

and 5mn – 2nl + 6lm = 0 …(2)

From (1), m = -(51 + 3n)

Putting the value of m in equation (2), we get,

-5(5l + 3n)n – 2nl – 6l(5l + 3n) = 0

∴ -25ln – 15n

^{2}– 2nl – 30l

^{2}– 18ln = 0

∴ – 30l

^{2}– 45ln – 15n

^{2}= 0

∴ 2l

^{2}+ 3ln + n

^{2}= 0

∴ 2l

^{2}+ 2ln + ln + n

^{2}= 0

∴ 2l(l + n) + n(l + n) = 0

∴ (l + n)(2l + n) = 0

∴ l + n = 0 or 2l + n = 0

l = -n or n = -2l

Now, m = -(5l + 3n), therefore, if l = -n,

m = -(-5n + 3n) = 2n

∴ the direction ratios of the second line are

a

_{2}= -1, b

_{2}= 1, c

_{2}= -2

Let θ be the angle between the lines.