Let vector equation of a plane be \(\vec{r}\cdot\left(2\hat{\imath}+\hat{\jmath}+2\hat{k}\right)=24\).
(A) The distance of the plane from the origin is 24.
(B) The distance of the plane from the origin is 0.
(C) The direction cosines of normal to the plane are <2, 1, 2>.
(D) The direction cosines of normal to the plane are \(\left\langle\displaystyle\frac{2}{3},\frac{1}{3},\frac{2}{3}\right\rangle\).
(E) The distance of the plane from the origin is 8.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (D) and (E) only
The normal \(\vec{n} = 2\hat{\imath}+\hat{\jmath}+2\hat{k}\) has length
\(|\vec{n}| = \sqrt{4+1+4} = 3\)
Dividing the equation by 3 gives the normal form \(\vec{r}\cdot\hat{n} = d\):
\( \displaystyle \vec{r}\cdot\left(\frac{2}{3}\hat{\imath}+\frac{1}{3}\hat{\jmath}+\frac{2}{3}\hat{k}\right) = 8 \)
So the distance of the plane from the origin is 8, and the direction cosines of the normal are \(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\).
Checking the statements
(A) The distance of the plane from the origin is 24. — Incorrect. 24 must be divided by \(|\vec{n}| = 3\); the distance is 8.
(B) The distance of the plane from the origin is 0. — Incorrect. The origin gives 0 ≠ 24, so the plane does not pass through it; the distance is 8.
(C) The direction cosines of normal to the plane are <2, 1, 2>. — Incorrect. These are direction ratios; \(2^2+1^2+2^2 = 9 \ne 1\).
(D) The direction cosines of normal to the plane are \(\left\langle\frac{2}{3},\frac{1}{3},\frac{2}{3}\right\rangle\). — Correct. Each ratio divided by \(|\vec{n}| = 3\).
(E) The distance of the plane from the origin is 8. — Correct. \(\frac{24}{3} = 8\).
Note: This topic was not in the CUET 2026 syllabus.