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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
The vertices $B$ and $C$ of a triangle $ABC$ lie on the line
$\frac{x}{1} = \frac{1 - y}{-2} = \frac{z - 2}{3}$.
The coordinates of $A$ and $B$ are $(1, 6, 3)$ and $(4, 9, \alpha)$ respectively and $C$ is at a distance of $10$ units from $B$. The area (in sq. units) of $\triangle ABC$ is:
Solution
$\frac{4}{1} = \frac{9-1}{2} = \frac{\alpha - 2}{3} \Rightarrow \alpha = 14$
Let $C(\lambda, 2\lambda + 1, 3\lambda + 2)$
$\vec{AD}\cdot(1\hat{i} + 2\hat{j} + 3\hat{k}) = 0$
$(\lambda - 1)\hat{i} + (2\lambda - 5)\hat{j} + (3\lambda - 1)\hat{k} \cdot (1\hat{i} + 2\hat{j} + 3\hat{k}) = 0$
$\Rightarrow \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$
$\Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$
$D = (1, 3, 5)$
$AD = \sqrt{3^2 + 2^2} = \sqrt{13}$
Area $(\triangle ABC) = \frac{1}{2} \times \sqrt{13} \times 10 = 5\sqrt{13}$
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