JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let a straight line \(L\) pass through the point \(P(2,-1,3)\) and be perpendicular to the lines \(\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2}\) and \(\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}\). If the line \(L\) intersects the \(y z\)-plane at the point \(Q\), then the distance between the points \(P\) and \(Q\) is :
- A \(\sqrt{10}\)
- B \(2 \sqrt{3}\)
- C 2
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
Vector parallel to \(L\) \(\begin{aligned} & =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{array}\right|=10 \hat{i}-10 \hat{j}+5 \hat{k} \\ & =5(2 \hat{i}-2 \hat{j}+\hat{k}) \end{aligned}\) Equation of ' \(L\) '…
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