The angle at which the normal to the plane 4x + 8y + z = 7 is inclined to y-axis is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\cos^{-1}\left(\displaystyle\frac{8}{9}\right)\)
The coefficients of x, y, z in the plane ax + by + cz = d are the direction ratios of its normal. Here they are 4, 8, 1.
\(\sqrt{4^2 + 8^2 + 1^2} = \sqrt{16 + 64 + 1} = \sqrt{81} = 9\)
So the direction cosines of the normal are
\( \displaystyle \frac{4}{9},\ \frac{8}{9},\ \frac{1}{9} \)
The second direction cosine is the cosine of the angle with the y-axis:
\( \displaystyle \cos\beta = \frac{8}{9},\quad \beta = \cos^{-1}\left(\frac{8}{9}\right) \)
Why the other options are incorrect
Option (1) \(\cos^{-1}\left(\displaystyle\frac{4}{9}\right)\) — Incorrect. This is the angle of the normal with the x-axis (coefficient of x).
Option (2) \(\cos^{-1}\left(\displaystyle\frac{1}{9}\right)\) — Incorrect. This is the angle of the normal with the z-axis (coefficient of z).
Option (4) \(\cos^{-1}\left(\displaystyle\frac{5}{9}\right)\) — Incorrect. 5/9 is not a direction cosine of the normal; the three are 4/9, 8/9 and 1/9.
Note: This topic was not in the CUET 2026 syllabus.