Chapter 5 Vectors Ex 5.3 Solutions

Let = =

and are perpendicular to each other. ∴ LHS = RHS Hence, ( + )2 = ( – )2.

Let = and = Consider ∙ = = (xc)(x) + (-6)(2) + (3)(2cx) = cx2 – 12 + 6cx = cx2 + 6cx – 12 If the angle between and is obtuse, ∙ < 0. ∴ cx2 + 6cx – 12 < 0 ∴ cx2 + 6cx < 12 ∴ c < 0. Hence, the angle between a and b is obtuse if c < 0.

Let = Projection of on X-axis Similarly, projections of on Y- and Z-axes are 3 and 4 respectively. ∴ sum of these projections = 2 + 3 + 4 = 9.

∵ , are non-zero vectors ∴ is perpendicular to Hence, the diagonals are perpendicular.

= = -3 = ∴ = i.e. is a non-zero scalar multiple of Hence, is parallel to .

The position vectors , and of the points P(0, -1, -2), Q(3, 1, 4) and R(5, 7, 1) are ∴ P = 45°

Let the triangle be denoted by ABC, where = , = and = ∵ , , are unit vectors. ∴ l(AB) = l(BC) = l(CA) = 1 ∴ the triangle is equilateral ∴ ∠A = ∠B = ∠C = 60° (i) Using the formula for angle between two vectors,

Let seg AB be a diameter of a circle with centre C and P be any point on the circle other than A and B. Then ∠APB is an angle subtended on a semicircle. Let = = and = . Then|| = || …(1) Hence, the angle subtended on a semicircle is the right angle.

Let α = 45°, β = 60° We have to find γ. ∴ cos2α + cos2β + cos2γ = 1 ∴ cos245° + cos260° + cos2γ = 1 Hence, the third direction angle is or .

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 m = cos 135° = cos (180° – 45°) = -cos 45° = and n = cos 45° = ∴ the direction cosines of the line are 0, .

The direction ratios of the line are a = 4, b = -12, c = 18. Let l, m, n be the direction cosines of the line.

The direction ratio of are -2, 2, 1. ∴ the direction cosines of are The coordinates of the points which are at a distance of d units from the point (x1, y1, z1) are given by (x1 ± ld, y1 ± md, z1 ± nd) Here x1 = 4, y1 = 1, z1 = 5, d = 6, l = , m = , n = ∴ the coordinates of the requited points are (4 ± 6, 1 ± (6), 5 ± (6)) i.e. (4 – 4, 1 + 4, 5 + 2) and (4 + 4, 1 – 4, 5 – 2) i.e. (0, 5, 7) and (8, -3, 3).

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 – 15n2 – 2nl – 30l2 – 18ln = 0 ∴ – 30l2 – 45ln – 15n2 = 0 ∴ 2l2 + 3ln + n2 = 0 ∴ 2l2 + 2ln + ln + n2 = 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 ∴ -l = = n ∴ ∴ the direction ratios of the first line are a1 = -1, b1 = 2, c1 = 1 If n = -2l, m = -(5l – 6l) – l ∴ l = m = ∴ ∴ the direction ratios of the second line are a2 = -1, b2 = 1, c2 = -2 Let θ be the angle between the lines.