JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the foot of the perpendicular from the point \(A (-1,4,3)\) on the plane \(P : 2 x + my + nz =4\), is \(\left(-2, \frac{7}{2}, \frac{3}{2}\right)\), then the distance of the point \(A\) from the plane \(P\), measured parallel to a line with direction ratios \(3,-1,-4\), is equal to.
- A \(1\)
- B \(\sqrt{26}\)
- C \(2 \sqrt{2}\)
- D \(\sqrt{14}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{26}\)
Step-by-step Solution
Detailed explanation
Let \(B\) be foot of \(\perp\) coordinates of \(B =\left(-2, \frac{7}{2}, \frac{3}{2}\right)\) Direction ratio of line \(AB\) is \(<2,1,3>\) so \(m=1, n=3\) So equation of \(AC\) is \(\frac{x+1}{3}=\frac{y-4}{-1}=\frac{z-3}{-4}=\lambda\) So point \(C\) is…
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