JEE Mains · Maths · STD 12 - 10. vector algebra
Let PQR be a triangle such that \(\overrightarrow{PQ}=-2\hat{i}-\hat{j}+2\hat{k}\) and \(\overrightarrow{PR}=a\hat{i}+b\hat{j}-4\hat{k}\), \(a, b\in\mathbb{Z}.\) Let S be the point on QR, which is equidistant from the lines PQ and PR. If \({|\overrightarrow{PR}|}=9\) and \(\overrightarrow{PS}=\hat{i}-7\hat{j}+2\hat{k}\), then the value of \(3a-4b\) is ___ .
- A 30
- B 37
- C 40
- D 35
Answer & Solution
Correct Answer
(B) 37
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{PQ}}=-2 \hat{i}-\hat{j}+2 \hat{k}\) \(\overrightarrow{\mathrm{PR}}=a \hat{\mathrm{i}}+b \hat{\mathrm{j}}-4 \hat{\mathrm{k}} \quad(a, b \in Z)\) \(\overrightarrow{\mathrm{PS}}=\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)…
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