**Chapter 5 Vectors Ex 5.4**

## Chapter 5 Vectors Ex 5.4

**Question 1.**

Solution:

Solution:

**Question 2.**

**Question 3.**

∴ θ = 60°.

**Question 4.**

∴ unit vectors perpendicular to both the vectors a¯ and b¯.

**Question 6.Solution:**

**Question 7.**

**Question 8.**

**Question 9.**

**Question 10.Find the area of parallelogram whose diagonals are determined by the **

**Question 11.**

**Question 12.**

By equality of vectors,

z – y = 0 ….(2)

x – z = 1 ……(3)

y – x = -1 ……(4)

From (2), y = z.

From (3), x = 1 + z

Substituting these values of x and y in (1), we get

**Question 13.Solution:**

By equality of vectors

2x = 0 i.e. x = 0

2y + z – 5 = 0 … (1)

2z – y = 0 … (2)

From (2), y = 2z

Substituting y = 2z in (1), we get

4z + z = 5 ∴ z = 1

∴ y = 2z = 2(1) = 2

∴ x = 0, y = 2, z = 1

**Question 15.Prove by vector method that sin (α + β) = sinα∙cosβ+cosα∙sinβ.Solution:**

Let ∠XOP and ∠XOQ be in standard position and m∠XOP = -α, m∠XOQ = β.

Take a point A on ray OP and a point B on ray OQ such that

OA = OB = 1.

Since cos (-α) = cos α

and sin (-α) = -sin α,

A is (cos (-α), sin (-α)),

i.e. (cos α, – sin α)

B is (cos β, sin β)

**Question 16.Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are(i) -2, 1, -1 and -3, -4, 1Solution:**

Let a, b, c be the direction ratios of the vector which is perpendicular to the two lines whose direction ratios are -2, 1, -1 and -3, -4, 1

∴ -2a + b – c = 0 and -3a – 4b + c = 0

Hence, the required direction ratios are -3, 5, 11.

**(ii) 1, 3, 2 and -1, 1, 2Solution:**

**Question 17.Solution:**

Given, al + bm + cn = 0 …(1)

and fmn + gnl + hlm = 0 …..(2)

Substituting this value of n in equation (2), we get

(fm + gl)∙[latex]-\left(\frac{a l+b m}{c}\right)[/latex] + hlm = 0

∴ -(aflm + bfm

^{2}+ agl

^{2}+ bglm) + chlm = 0

∴ agl

^{2}+ (af + bg – ch)lm + bfm

^{2}= 0 … (4)

Note that both l and m cannot be zero, because if l = m = 0, then from (3), we get

n = 0, which is not possible as l

^{2}+ m

^{2}+ n

^{2}= 1.

Let us take m # 0.

Dividing equation (4) by m

^{2}, we get

**Question 18.If A(1, 2, 3) and B(4, 5, 6) are two points, then find the foot of the perpendicular from the point B to the line joining the origin and point A.Solution:**

Let M be the foot of the perpendicular drawn from B to the line joining O and A.

Let M = (x, y, z)

OM has direction ratios x – 0, y – 0, z – 0 = x, y, z

OA has direction ratios 1 – 0, 2 – 0, 3 – 0 = 1, 2, 3

But O, M, A are collinear.

∴ x = k, y = 2k, z = 3k

∴ M = (k, 2k, 3k)

∵ BM has direction ratios

k – 4, 2k – 5, 3k – 6

BM is perpendicular to OA

∴ (l)(k – 4) + 2(2k – 5) + 3(3k – 6)

∴ = k – 4 + 4k – 10 + 9k – 18 = 0

∴ 14k = 32