Question:
The distance of the point having position vector \(-\hat{i}+2\hat{j}+6\hat{k}\) from the straight line passing through the point \((2,3,-4)\) and parallel to the vector \(6\hat{i}+3\hat{j}-4\hat{k}\) is :
The distance of the point having position vector \(-\hat{i}+2\hat{j}+6\hat{k}\) from the straight line passing through the point \((2,3,-4)\) and parallel to the vector \(6\hat{i}+3\hat{j}-4\hat{k}\) is :
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Distance of a point \(P\) from a line through \(A\) with direction \(\vec d\) is \(|\vec{AP}\times\vec d|/|\vec d|\).
Updated On: Nov 5, 2026
- \(6\)
- \(7\)
- \(2\sqrt{13}\)
- \(4\sqrt{3}\)
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The Correct Option is B
Solution and Explanation
Let \(P(-1,2,6)\), \(A(2,3,-4)\), \(\vec d=6\hat i+3\hat j-4\hat k\). Then \(\vec{AP}=-3\hat i-\hat j+10\hat k\).
\(\vec{AP}\times\vec d=\begin{vmatrix}\hat i&\hat j&\hat k\\-3&-1&10\\6&3&-4\end{vmatrix}=\hat i(4-30)-\hat j(12-60)+\hat k(-9+6)=-26\hat i+48\hat j-3\hat k\).
\(|\vec{AP}\times\vec d|^2=676+2304+9=2989\), \(|\vec d|^2=36+9+16=61\).
Distance \(=\sqrt{2989/61}=\sqrt{49}=7\).
Answer: (2) \(7\)
\(\vec{AP}\times\vec d=\begin{vmatrix}\hat i&\hat j&\hat k\\-3&-1&10\\6&3&-4\end{vmatrix}=\hat i(4-30)-\hat j(12-60)+\hat k(-9+6)=-26\hat i+48\hat j-3\hat k\).
\(|\vec{AP}\times\vec d|^2=676+2304+9=2989\), \(|\vec d|^2=36+9+16=61\).
Distance \(=\sqrt{2989/61}=\sqrt{49}=7\).
Answer: (2) \(7\)
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