The angle between the line \(\displaystyle\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}\) and the plane \(10x+2y-11z=3\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\sin^{-1}\left(\displaystyle\frac{8}{21}\right)\)
If \(\vec{b}\) is along the line and \(\vec{n}\) is normal to the plane, the angle \(\theta\) between the line and the plane satisfies
\( \displaystyle \sin\theta = \frac{\left|\vec{b}\cdot\vec{n}\right|}{|\vec{b}|\,|\vec{n}|} \)
Here
\(\vec{b} = 2\hat{\imath} + 3\hat{\jmath} + 6\hat{k}\)
\(\vec{n} = 10\hat{\imath} + 2\hat{\jmath} - 11\hat{k}\)
\(\vec{b}\cdot\vec{n} = 20 + 6 - 66 = -40\)
\(|\vec{b}| = \sqrt{4 + 9 + 36} = 7\)
\(|\vec{n}| = \sqrt{100 + 4 + 121} = 15\)
So
\( \displaystyle \sin\theta = \frac{40}{7 \times 15} = \frac{40}{105} = \frac{8}{21} \)
\( \displaystyle \theta = \sin^{-1}\left(\frac{8}{21}\right) \)
Why the other options are incorrect
Option (1) \(\sin^{-1}\left(-\displaystyle\frac{8}{21}\right)\) — Incorrect. This leaves out the modulus of \(\vec{b}\cdot\vec{n} = -40\); the angle between a line and a plane is taken between 0 and \(\frac{\pi}{2}\), so its sine is not negative.
Option (2) \(\sin^{-1}\left(-\displaystyle\frac{21}{8}\right)\) — Incorrect. A sine value cannot be less than −1, so this angle does not exist.
Option (3) \(\sin^{-1}\left(\displaystyle\frac{21}{8}\right)\) — Incorrect. A sine value cannot be more than 1, so this angle does not exist.
Note: This topic was not in the CUET 2026 syllabus.