Question

The acute angle between the lines
x = −2 + 2t, y = 3 − 4t, z = −4 + t and
x = −2 − t, y = 3 + 2t, z = −4 + 3t is

पर्याय

• \[\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)\]

• \[\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)\]

• \[\cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\]

• \[\cos^{1}\left(\frac{2}{\sqrt{6}}\right)\]

Answer 1

\[\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)\]

Explanation:

Given equations of lines are
x = −2 + 2t, y = 3 − 4t, z = −4 + t

\[\therefore\quad\frac{x+2}{2}=\frac{y-3}{-4}=\frac{z+4}{1}=\mathrm{t}\]

and

x = −2 − t, y = 3 + 2t, z = −4 + 3t

\[\therefore\quad\frac{x+2}{-1}=\frac{y-3}{2}=\frac{z+4}{3}=\mathrm{t}\]

a₁, b₁, c₁ = 2, −4, 1
a₂, b₂, c₂ = −1, 2, 3

\[\cos\theta=\left|\frac{\mathrm{a}_1\mathrm{a}_2+\mathrm{b}_1\mathrm{b}_2+\mathrm{c}_1\mathrm{c}_2}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2+\mathrm{c}_1^2}\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}}\right|\]

\[=\left|\frac{-2-8+3}{\sqrt{4+16+1}\sqrt{1+4+9}}\right|\]

\[=\left|\frac{7}{\sqrt{21}\sqrt{14}}\right|\]

\[=\frac{1}{\sqrt{6}}\]

\[\therefore\quad\theta=\cos^{-1}\left(\frac{1}{\sqrt{6}}\right)\]