Vectors — Maths STD 12 Science — Question

$\begin{aligned} & \Rightarrow-30 n^2-45 n l-\left.15\right|^2=0 \\ & \Rightarrow 2 n^2+3 n l+\left.\right|^2=0 \\ & \Rightarrow 2 n^2+2 n l+n l+\left.\right|^2=0\end{aligned}$

⇒ (2n + l) (n + l) = 0 ∴ 2n + l = 0 OR n + l = 0 ∴ l = -2n OR l = -n ∴ l = -2n From (2), 3l + m + 5n = 0 ∴ -6n + m + 5n = 0 ∴ m = n i.e. (-2n, n, n) = (-2, 1, 1) ∴ l = -n ∴ -3n + m + 5n = 0 ∴ m = -2n i.e. (-n, -2n, n) = (1, 2, -1) (a1, b1, c1) = (-2, 1, 1) and (a2, b3, c3) = (1, 2, -1)

$\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \cdot \sqrt{a_2^2+b^2-2+c_2^2}}\right|$

$=\left|\frac{(2)(1)+(-1)(2)+(-1)(-1)}{\sqrt{(2)^2+1^2+1^2} \cdot \sqrt{1^2+2^2+(1)^2}}\right|$

$=\left|\frac{2-2+1}{\sqrt{6} \cdot \sqrt{6}}\right|$

$=\left|-\frac{1}{6}\right|=\frac{1}{6}$

$\theta=\cos ^{-1}\left(\frac{1}{6}\right)$