JEE Mains · Maths · STD 12 - 11. three dimension geometry
The vertices \(B\) and \(C\) of a \(\Delta ABC\) lie on the line, \(\frac{{x + 2}}{3} = \frac{{y - 1}}{0} = \frac{z}{4}\) such that \(BC = 5\, units\). Then the area (in \(sq. units\)) of this triangle, given that the point \(A\, (1, -1, 2)\) is
- A \(2\sqrt {34} \)
- B \(\sqrt {34} \)
- C \(6\)
- D \(5\sqrt {17} \)
Answer & Solution
Correct Answer
(B) \(\sqrt {34} \)
Step-by-step Solution
Detailed explanation
Let any point on given line is \(D(3 \lambda-2,1,4 \lambda)\) Now \(A D \perp B C\) \(\mathrm{D} \mathrm{R}\) of \(BC\) \(\Rightarrow a_{1}=3, b_{1}=0, c_{1}=y\) \(D \cdot R,\) of \(A D\) \(\Rightarrow a_{2}=3 \lambda-3, b_{2}=2, c_{2}=4 \lambda-2\)…
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