The distance of the plane \(\vec{r}\cdot(2\hat{\imath}-\hat{\jmath}+\hat{k})=3\) from the origin is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{\sqrt{6}}\)
For the plane \(\vec{r}\cdot\vec{n} = d\), the distance from the origin is \(\frac{|d|}{|\vec{n}|}\). Here \(\vec{n} = 2\hat{\imath}-\hat{\jmath}+\hat{k}\) and d = 3:
\(|\vec{n}| = \sqrt{4 + 1 + 1} = \sqrt{6}\)
\( \displaystyle \text{Distance} = \frac{3}{\sqrt{6}} \)
Note: This topic was not in the CUET 2026 syllabus.