CUET · MATHS · PYQ PAPER 2023
The distance between the line \(\vec{r}=2 \hat{i}+2 \hat{j}-3 \hat{k}+\lambda(\hat{i}+\hat{j}+4 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}-5 \hat{j}+\hat{k})-4=0\) is:
- A \(0\)
- B \(2 \sqrt{3}\)
- C \(\frac{4 \sqrt{3}}{3}\)
- D \(\frac{5 \sqrt{3}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{5 \sqrt{3}}{3}\)
Step-by-step Solution
Detailed explanation
\(\vec{b} \cdot \vec{n} = (\hat{i}+\hat{j}+4\hat{k}) \cdot (\hat{i}-5\hat{j}+\hat{k}) = 1-5+4 = 0\) Distance \(D = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}\) \(D = \frac{|(2\hat{i}+2\hat{j}-3\hat{k}) \cdot (\hat{i}-5\hat{j}+\hat{k}) - 4|}{|\hat{i}-5\hat{j}+\hat{k}|}\)…
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