Let P be the point (10, –2, –1) and Q be the foot of the perpendicular drawn from the point R(1, 7, 6) on the line passing through the points (2, –5, 11) and (–6, 7, –5). Then the length of the line segment PQ is equal to __________. [2024]

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Solution

Q.
Let P be the point (10, –2, –1) and Q be the foot of the perpendicular drawn from the point R(1, 7, 6) on the line passing through the points (2, –5, 11) and (–6, 7, –5). Then the length of the line segment PQ is equal to __________. [2024]

Ans.
(13)

Equation of line passing through (2, –5, 11) and (–6, 7, –5) is

$L : \frac{x + 6}{8} = \frac{y - 7}{- 12} = \frac{z + 5}{16}$

i.e., $\frac{x + 6}{2} = \frac{y - 7}{- 3} = \frac{z + 5}{4} = λ$ ... (i)

$\Rightarrow$ any point Q on the given line is $(2 λ - 6, - 3 λ + 7, 4 λ - 5)$

Now, D.r.'s of QR = $(2 λ - 7, - 3 λ, 4 λ - 11)$

Since QR is perpendicular to L

$\Rightarrow 2 (2 λ - 7) - 3 (- 3 λ) + 4 (4 λ - 11) = 0$

$\Rightarrow 29 λ - 58 = 0 \Rightarrow λ = 2$

So Q(–2, 1, 3) is the required point

and $| \vec{P Q} |$ $= \sqrt{{(- 2 - 10)}^{2} + {(1 + 2)}^{2} + {(3 + 1)}^{2}}$ = $\sqrt{169} = 13$ .

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