JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(Q\) be the foot of the perpendicular from the point \(\mathrm{P}(7,-2,13)\) on the plane containing the lines \(\frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}\) and \(\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}\) Then \((\mathrm{PQ})^{2}\), is equal to ..... .
- A \(100\)
- B \(96\)
- C \(97\)
- D \(95\)
Answer & Solution
Correct Answer
(B) \(96\)
Step-by-step Solution
Detailed explanation
Containing the line \(\left|\begin{array}{ccc}x+1 & y-1 & z-3 \\ 6 & 7 & 8 \\ 3 & 5 & 7\end{array}\right|=0\) \(9(x+1)-18(y-1)+9(z-3)=0\) \(x-2 y+z=0\) \(P Q=\left|\frac{7+4+13}{\sqrt{6}}\right|=4 \sqrt{6}\) \(P Q^{2}=96\)
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