CUET · MATHS · PYQ PAPER 2023
The distance of the plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=3\) from the origin is :
- A \(\frac{3}{\sqrt{6}}\)
- B \(\frac{\sqrt{3}}{2}\)
- C \(\frac{8}{\sqrt{6}}\)
- D \(\frac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{\sqrt{6}}\)
Step-by-step Solution
Detailed explanation
\( \vec{n} = 2\hat{i}-\hat{j}+\hat{k}, d=3 \) \( \|\vec{n}\| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{4+1+1} = \sqrt{6} \) \( \text{Distance} = \frac{|d|}{\|\vec{n}\|} = \frac{3}{\sqrt{6}} \)
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