Q.If the line $\dfrac{x}{3}=\dfrac{y}{4}=z$ is perpendicular to the line $\dfrac{x-1}{k}=\dfrac{y+2}{3}=\dfrac{z-3}{k-1}$ then the value of $k$ is:
(A) $\dfrac{11}{4}$
(B) $-\dfrac{11}{4}$
(C) $\dfrac{11}{2}$
(D) $\dfrac{4}{11}$

Use $\cos\theta = \dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$.

Direction ratios: $(a_1,b_1,c_1) = (3,2,6)$ and $(a_2,b_2,c_2) = (-2,1,2)$.

$\cos\theta = \frac{3(-2)+2(1)+6(2)}{\sqrt{3^2+2^2+6^2}\,\sqrt{(-2)^2+1^2+2^2}} = \frac{-6+2+12}{\sqrt{9+4+36}\,\sqrt{4+1+4}} = \frac{8}{\sqrt{49}\sqrt{9}} = \frac{8}{7\times3} = \frac{8}{21}$

Since we want the acute angle, we take the positive value:

$\theta = \cos^{-1}\left(\frac{8}{21}\right)$

This matches option (d).

Two lines are at right angles when the dot product of their direction ratios is zero.

Direction ratios of the lines:

Line 1: $(-3,\,2k,\,2)$

Line 2: $(3k,\,1,\,-5)$

For lines at right angles, the dot product of direction ratios must be zero:

$(-3)(3k)+(2k)(1)+(2)(-5)=0$

$-9k+2k-10=0$

$-7k=10$

$k=-\frac{10}{7}$