Let A and B be two distinct points on the line . Both A and B are at a distance from the foot of perpendicular drawn from the point (1, 2, 3) on the line L. If O is the origin, then is equal to [2025]

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Solution

Q.
Let A and B be two distinct points on the line $L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Both A and B are at a distance $2 \sqrt{17}$ from the foot of perpendicular drawn from the point (1, 2, 3) on the line L. If O is the origin, then $\vec{O A} \cdot \vec{O B}$ is equal to [2025]

1 49

2 62

3 21

4 47

Ans.
(4)

We have, $L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2} = λ (say)$

$\Rightarrow$ Let foot of perpendicular from P(1, 2, 3) on L is

$Q = (3 λ + 6, 2 λ + 7, - 2 λ + 7)$

Now, $3 (3 λ + 6 - 1) + 2 (2 λ + 7 - 2) - 2 (- 2 λ + 7 - 3) = 0$ [ $∵ \vec{P Q} ⊥ L$ ]

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

Now, distance of A from Q(3, 5, 9) is the foot of perpendicular.

Let any point on line L is $A (3 μ + 6, 2 μ + 7, - 2 μ + 7)$

Now, distance of A from Q is $2 \sqrt{17}$

$\Rightarrow {(3 μ + 6 - 3)}^{2} + {(2 μ + 7 - 5)}^{2} + {(- 2 μ + 7 - 9)}^{2} = {(2 \sqrt{17})}^{2}$

$\Rightarrow 9 μ^{2} + 9 + 18 μ + {4 μ^{2} + 4 + 8 μ + 4 μ}^{2} + 4 + 8 μ = 68$

$\Rightarrow 17 μ^{2} + 17 + 34 μ = 68 \Rightarrow μ^{2} + 2 μ + 1 = 4$

$\Rightarrow μ^{2} + 2 μ - 3 = 0 \Rightarrow (μ + 3) (μ - 1) = 0 \Rightarrow μ = - 3 or 1$

$∴ A (- 3, 1, 13) and B (9, 9, 5)$ are the points on the line L.

$∴ \vec{O A} \cdot \vec{O B} = - 27 + 9 + 65 = 47$ .

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