The vertices B and C of a triangle ABC lie on the line The coordinates of A and B are (1,6,3) and (4,9,) respectively and C is at a distance of 10 units from B. The area (in sq. units) of ABC is: [2026]

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Solution

Q.
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, $α$ ) respectively and C is at a distance of 10 units from B. The area (in sq. units) of $∆$ ABC is: [2026]

1 $5 \sqrt{13}$

2 $10 \sqrt{13}$

3 $20 \sqrt{13}$

4 $15 \sqrt{13}$

Ans.
(1)

$\frac{4}{1} = \frac{9 - 1}{2} = \frac{α - 2}{3} \Rightarrow α = 14$

$\vec{A D} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 0$

$(λ - 1) \hat{i} + (2 λ - 5) \hat{j} + (3 λ - 1) \hat{k} = \vec{A D}$

$\Rightarrow λ - 1 + 4 λ - 10 + 9 λ - 3 = 0$

$\Rightarrow 14 λ = 14 \Rightarrow λ = 1$

$D = (1, 3, 5)$

$A D = \sqrt{3^{2} + 2^{2}} = \sqrt{13}$

$Ar (∆ A B C) = \frac{1}{2} \times \sqrt{13} \times 10 = 5 \sqrt{13}$

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