CUET · MATHS · PYQ PAPER 2023
The angle between the line \(\vec{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\) and the plane \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=4\) is :
- A \(\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\)
- B \(\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\)
- C \(\tan ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\)
- D \(\sin ^{-1}\left(\frac{2 \sqrt{3}}{5}\right)\)
Answer & Solution
Correct Answer
(A) \(\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\)
Step-by-step Solution
Detailed explanation
\(\vec{b} = \hat{i}-\hat{j}+\hat{k}\) \(\vec{n} = 2 \hat{i}-\hat{j}+\hat{k}\) \(\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}\) \(\sin \theta = \frac{|(1)(2)+(-1)(-1)+(1)(1)|}{\sqrt{1^2+(-1)^2+1^2} \sqrt{2^2+(-1)^2+1^2}}\)…
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