Chapter 5 Vectors Ex 5.4 Solutions

Let = , =

Let θ be the angle between and ∴ θ = 60°.

Given : = and = ∴ unit vectors perpendicular to both the vectors and . ∴ required vectors of magnitude 5 units = ±

Let θ be the angle between and . Then = 8 gives sin θ = 8 ∴ 2 × 5 × sin θ = 8

Given : = , = , =

Given : = , = Area of the parallelogram whose adjacent sides are and is = sq units.

Let ABCD be a parallelogram. Then = +

Given: = , =

, , and are four distinct vectors

Given = , = 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 1 + z + z + z = 3 ∴ z = ∴ y = z = ∴ x = 1 + z =1 +

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 ∴

Let θ be the angle between and . Then = gives Hence, the angle between and is .

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) The angle between and is α + β. Also , lie in the XY-plane. ∴ the unit vector perpendicular to and is . ∴ × = [OA∙OB sin (α + β)] = sin(α + β)∙ …(2) ∴ from (1) and (2), sin (α + β) = sin α cos β + cos α sin β.

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 ∴ the required direction ratios are -3, 5, 11 Alternative Method: Let and be the vectors along the lines whose direction ratios are -2, 1, -1 and -3, -4, 1 respectively. Then = and = The vector perpendicular to both and is given by Hence, the required direction ratios are -3, 5, 11.

Given, al + bm + cn = 0 …(1) and fmn + gnl + hlm = 0 …..(2) From (1), n = …..(3) Substituting this value of n in equation (2), we get (fm + gl)∙-≤ft( + hlm = 0 ∴ -(aflm + bfm2 + agl2 + bglm) + chlm = 0 ∴ agl2 + (af + bg – ch)lm + bfm2 = 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 l2 + m2 + n2 = 1. Let us take m # 0. Dividing equation (4) by m2, we get ag + (af + bg – ch) + bf = 0 … (5) This is quadratic equation in . If l1, m1, n1 and l2, m2, n2 are the direction cosines of the two lines given by the equation (1) and (2), then and are the roots of the equation (5). From the quadratic equation (5), we get i.e. if = 0.

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. ∴ = k …(Let) ∴ 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 ∴ k = ∴ M = (k, 2k, 3k) = ()