MHT CET · Maths · Three Dimensional Geometry
The angle between the line \(\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}\) and the plane \(\bar{r} \cdot(6 \hat{\imath}-2 \hat{\jmath}-3 \hat{k})=5\)
is
- A \(\sin ^{-1}\left(\frac{4}{21}\right)\)
- B \(\cos ^{-1}\left(\frac{4}{21}\right)\)
- C \(\sin ^{-1}\left(\frac{5}{7}\right)\)
- D \(\cos ^{-1}\left(\frac{5}{7}\right)\)
Answer & Solution
Correct Answer
(A) \(\sin ^{-1}\left(\frac{4}{21}\right)\)
Step-by-step Solution
Detailed explanation
Angle between line and plane is given by
\(\sin \theta=\frac{a a_{1}+b b_{1}+c c_{1}}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}\)
Here \(a=2, b=1, c=2\) and \(a_{1}=6, b_{1}=-2, c_{1}=-3\)
\(\therefore \sin \theta=\frac{12-2-6}{3 \times 7}=\frac{4}{21} \Rightarrow \theta=\sin ^{-1}\left(\frac{4}{21}\right)\)
\(\sin \theta=\frac{a a_{1}+b b_{1}+c c_{1}}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}\)
Here \(a=2, b=1, c=2\) and \(a_{1}=6, b_{1}=-2, c_{1}=-3\)
\(\therefore \sin \theta=\frac{12-2-6}{3 \times 7}=\frac{4}{21} \Rightarrow \theta=\sin ^{-1}\left(\frac{4}{21}\right)\)
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