The angle between the lines \(\frac{{x - 4}}{2} = \frac{y}{1} = \frac{{z + 1}}{{ - 2}},\frac{{x - 1}}{4} = \frac{{y + 1}}{{ - 4}} = \frac{{z - 2}}{2} \) is:

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Question

This question was previously asked in

AAI ATC Junior Executive 27 July 2022 Shift 1 Official Paper

Answer 1

Concept:

The angle between the pair of lines

\(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \)

and \(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2} \) is given by

\(cosθ =\left|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2} }\sqrt{a_2^{2}+b_2^{2}+c_2^{2} }} \right| \)

Solution:

Given the pair of lines \(\frac{{x - 4}}{2} = \frac{y}{1} = \frac{{z + 1}}{{ - 2}}\) and

\(\frac{{x - 1}}{4} = \frac{{y + 1}}{{ - 4}} = \frac{{z - 2}}{2}\)

a₁ = 2, b₁ = 1, c₁ = -2

a₂ = 4, b₂= -4, c₂ = 2

\(cosθ =\left|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2} }\sqrt{a_2^{2}+b_2^{2}+c_2^{2} }} \right|\)

\(cosθ =\left|\frac{2\times 4-1\times 4-2\times 2}{\sqrt{2^{2}+1^{2}+(-2)^{2} }\sqrt{4^{2}+(-4)^{2}+2^{2} }} \right| \)

\(cosθ =\left|\frac{8-8}{\sqrt{9}\sqrt{36}} \right|\)

\(cosθ =\left|\frac{0}{3\times 6} \right|\)

cosθ = 0

θ = \(\frac{\pi}{2}\)