KCET · Maths · Three Dimensional Geometry
Acute angle between the line \( \frac{x-5}{2}=\frac{y+1}{-1}=\frac{z+4}{1} \) and the plane \( 3 x-4 y-z+5=0 \) is
- A \( \sin ^{-1}\left(\frac{9}{\sqrt{364}}\right) \)
- B \( \sin ^{-1}\left(\frac{9}{2 \sqrt{13}}\right) \)
- C \( \cos ^{-1}\left(\frac{9}{\sqrt{364}}\right) \)
- D \( \cos ^{-1}\left(\frac{5}{2 \sqrt{13}}\right) \)
Answer & Solution
Correct Answer
(D) \( \cos ^{-1}\left(\frac{5}{2 \sqrt{13}}\right) \)
Step-by-step Solution
Detailed explanation
\( L: \frac{(x-5)}{2}=\frac{(y+1)}{-1}=\frac{(z+4)}{1} \)
Dr's of line: \( 2,-1,1 \)
P: \( 3 x-4 y-z+5=0 \)
Dr's of normal to plane: \( -3,4,1 \)
\( \sin \theta=\left|\frac{-6-4+1}{\sqrt{4+1+1 \sqrt{9+16+1}}}\right|=\left|\frac{-9}{\sqrt{156}}\right| \)
\[
\begin{array}{l}
\sin \theta=\frac{9}{\sqrt{156}} \\
\theta=\sin ^{-1}\left(\frac{9}{\sqrt{156}}\right) \\
\theta=\sin ^{-1}\left(\frac{9}{2 \sqrt{39}}\right) \\
\theta=\cos ^{-1}\left(\frac{5}{2 \sqrt{13}}\right)
\end{array}
\]
Dr's of line: \( 2,-1,1 \)
P: \( 3 x-4 y-z+5=0 \)
Dr's of normal to plane: \( -3,4,1 \)
\( \sin \theta=\left|\frac{-6-4+1}{\sqrt{4+1+1 \sqrt{9+16+1}}}\right|=\left|\frac{-9}{\sqrt{156}}\right| \)
\[
\begin{array}{l}
\sin \theta=\frac{9}{\sqrt{156}} \\
\theta=\sin ^{-1}\left(\frac{9}{\sqrt{156}}\right) \\
\theta=\sin ^{-1}\left(\frac{9}{2 \sqrt{39}}\right) \\
\theta=\cos ^{-1}\left(\frac{5}{2 \sqrt{13}}\right)
\end{array}
\]
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