**Chapter 5 Vectors Ex 5.4**

## Chapter 5 Vectors Ex 5.4

**Question 1.**

Solution:

Solution:

**Question 2.**

**Question 3.**

âˆ´ Î¸ = 60Â°.

**Question 4.**

âˆ´ unit vectors perpendicular to both the vectors aÂ¯ and bÂ¯.

**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 â€“ y = 0 â€¦.(2)

x â€“ z = 1 â€¦â€¦(3)

y â€“ x = -1 â€¦â€¦(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 â€“ 5 = 0 â€¦ (1)

2z â€“ y = 0 â€¦ (2)

From (2), y = 2z

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

4z + z = 5 âˆ´ z = 1

âˆ´ y = 2z = 2(1) = 2

âˆ´ x = 0, y = 2, z = 1

**Question 15.Prove by vector method that sin (Î± + Î²) = sinÎ±âˆ™cosÎ²+cosÎ±âˆ™sinÎ².Solution:**

Let âˆ XOP and âˆ XOQ be in standard position and mâˆ XOP = -Î±, mâˆ XOQ = Î².

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

OA = OB = 1.

Since cos (-Î±) = cos Î±

and sin (-Î±) = -sin Î±,

A is (cos (-Î±), sin (-Î±)),

i.e. (cos Î±, â€“ sin Î±)

B is (cos Î², sin Î²)

**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

âˆ´ -2a + b â€“ c = 0 and -3a â€“ 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 â€¦(1)

and fmn + gnl + hlm = 0 â€¦..(2)

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

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

âˆ´ -(aflm + bfm

^{2}+ agl

^{2}+ bglm) + chlm = 0

âˆ´ agl

^{2}+ (af + bg â€“ ch)lm + bfm

^{2}= 0 â€¦ (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 â€“ 0, y â€“ 0, z â€“ 0 = x, y, z

OA has direction ratios 1 â€“ 0, 2 â€“ 0, 3 â€“ 0 = 1, 2, 3

But O, M, A are collinear.

âˆ´ x = k, y = 2k, z = 3k

âˆ´ M = (k, 2k, 3k)

âˆµ BM has direction ratios

k â€“ 4, 2k â€“ 5, 3k â€“ 6

BM is perpendicular to OA

âˆ´ (l)(k â€“ 4) + 2(2k â€“ 5) + 3(3k â€“ 6)

âˆ´ = k â€“ 4 + 4k â€“ 10 + 9k â€“ 18 = 0

âˆ´ 14k = 32