Let in a ABC, the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line . Then the area (in sq. units) of ABC is : [2025]

Question

Let in a $△$ ABC, the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Then the area (in sq. units) of $△$ ABC is : [2025]

Smart Achievers · Accepted Answer

(4)

Let BM be the height of the triangle ABC.

Direction ratios of AC = 3, 2, –2

Coordinates of $M = (3 λ + 6, 2 λ + 7, - 2 λ + 7)$

Direction ratios of BM $= (3 λ + 6 - 1, 2 λ + 7 - 2, - 2 λ + 7 - 3)$

$= (3 λ + 5, 2 λ + 5, - 2 λ + 4)$

$∵ B M ⊥ A C$

$∴ 3 (3 λ + 5) + 2 (2 λ + 5) - 2 (- 2 λ + 4) = 0$

$\Rightarrow 9 λ + 15 + 4 λ + 10 + 4 λ - 8 = 0$

$\Rightarrow 17 λ + 17 = 0 \Rightarrow λ = - 1$

$∴$ Coordinates of M = (3, 5, 9)

$∴ B M = \sqrt{{(3 - 1)}^{2} + {(5 - 2)}^{2} + {(9 - 3)}^{2}} = 7$

Area of $△ A B C = = \frac{1}{2} \times 7 \times 6 = 21 sq . units$ .

Solution

Q.
Let in a $△$ ABC, the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Then the area (in sq. units) of $△$ ABC is : [2025]

1 56

2 17

3 42

4 21

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