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JEE Mains · Maths · STD 12 - 11. three dimension geometry

Let the line \(\mathrm{L}\) be the projection of the line  \(\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}\) in the plane \(x-2 y-z=3 .\) If \(d\) is the distance of the point \((0,0,6)\) from \(\mathrm{L}\), then \(\mathrm{d}^{2}\) is equal to .... .

  1. A \(48\)
  2. B \(26\)
  3. C \(14\)
  4. D \(1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(26\)

Step-by-step Solution

Detailed explanation

\(\mathrm{L}_{1}: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}\) for foot of \(\perp \mathrm{r}\) of \((1,3,4)\) on \(x-2 y-z-3=0\) \((1+t)-2(3-2 t)-(4-t)-3=0\) \(\Rightarrow \mathrm{t}=2\) So foot of \(\perp \mathrm{r} \triangleq(3,-1,2)\) and point of intersection of…
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