The lines x – 2 2 = y – 2 = z – 7 16 and x + 3 4 = y + 2 3 = z + 2 1 intersect at the point P. If the distance of P from the line x + 1 2 = y – 1 3 = z – 1 1 is l , then 14 l 2 is equal to _________. [2024]

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Solution

Q.
The lines $\frac{x - 2}{2} = \frac{y}{- 2} = \frac{z - 7}{16}$ and $\frac{x + 3}{4} = \frac{y + 2}{3} = \frac{z + 2}{1}$ intersect at the point P. If the distance of P from the line $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1}$ is $l$ , then $14 l^{2}$ is equal to _________. [2024]

Ans.
(108)

We have, $\frac{x - 2}{2} = \frac{y}{- 2} = \frac{z - 7}{16} = λ$ (say)

$\Rightarrow x = 2 λ + 2, y = - 2 λ, z = 16 λ + 7$

Also, $\frac{x + 3}{4} = \frac{y + 2}{3} = \frac{z + 2}{1} = μ$ (say)

$\Rightarrow x = 4 μ - 3, y = 3 μ - 2, z = μ - 2$

Lines are intersecting at point P.

$∴ 2 λ + 2 = 4 μ - 3 \Rightarrow 2 λ - 4 μ = - 5$ ... (i)

$- 2 λ = 3 μ - 2 \Rightarrow - 2 λ - 3 μ = - 2$ ... (ii)

On solving (i) and (ii), we get

$λ = - \frac{1}{2}$ and $μ = 1$

$∴$ Point P is (1, 1, –1)

Now, $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1} = k$ (say)

$\Rightarrow$ x = 2k – 1, y = 3k + 1, z = k + 1

D.r.'s of PQ : 2k – 2, 3k, k + 2

D.r.'s of line $\frac{x + 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{1}$ is 2, 3, 1

As both line are perpendicular to each other.

2(2k –2) + 3(3k) + 1(k +2) = 0

$\Rightarrow 14 k = 2 \Rightarrow k = \frac{1}{7}$

Thus, point Q is $(- \frac{5}{7}, \frac{10}{7}, \frac{8}{7})$

$∴ P Q = \sqrt{{(1 + \frac{5}{7})}^{2} + {(1 - \frac{10}{7})}^{2} + {(- 1 - \frac{8}{7})}^{2}} = \sqrt{\frac{378}{49}}$

Also, $P Q^{2} = l^{2} = \frac{378}{49} \Rightarrow 14 l^{2} = \frac{378}{49} \times 14 = 108$ .

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