Let the co-ordinates of one vertex of ∆ A B C be A ( 0 , 2 , α ) and the other two vertices lie on the line x + α 5 = y - 1 2 = z + 4 3 . For α ∈ ℤ , if the area of ∆ A B C is 21 sq . units and the li

Question

Let the co-ordinates of one vertex of ∆ A B C be A ( 0 , 2 , α ) and the other two vertices lie on the line x + α 5 = y - 1 2 = z + 4 3 . For α ∈ ℤ , if the area of ∆ A B C is 21 sq . units and the line segment B C has length 2 21 units, then α 2 is equal to _______.

TestHub · Accepted Answer

Given,

The co-ordinates of one vertex of $∆ A B C$ be $A (0, 2, α)$ and the other two vertices lie on the line $\frac{x + α}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$

Now plotting the diagram we get,

Now any point on the plane $\frac{x + α}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = k$ is given by $(5 k - α, 2 k + 1, 3 k - 4)$

Now direction ratio of line $A D$ is given by, $(5 k - α - 0, 2 k + 1 - 2, 3 k - 4 - α)$

$\equiv (5 k - α, 2 k - 1, 3 k - 4 - α)$

Now $A D$ is perpendicular to $B C$ ,

So, $5 (5 k - α) + 2 (2 k - 1) + 3 (3 k - 4 - α) = 0$

$\Rightarrow 19 k - 4 α - 7 = 0 . . . . . (1)$

Also given area, $21 = \frac{1}{2} \times 2 \sqrt{21} \times A D$

$\Rightarrow A D = \sqrt{21}$

$\Rightarrow (5 k - α) + (2 k - 1) + (3 k - 4 - α) = 21$

$\Rightarrow 19 k^{2} - 8 k α + α^{2} - 14 k + 4 α = 2 . . . . . . (2)$

Now on solving equation $(1) & (2)$ we get,

$α = 3 \Rightarrow α^{2} = 9$ .

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