Let a straight line L pass through the point P(2, –1, 3) and be perpendicular to the lines and . If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is : [2025]

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Solution

Q.
Let a straight line L pass through the point P(2, –1, 3) and be perpendicular to the lines $\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2}$ and $\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}$ . If the line L intersects the yz-plane at the point Q, then the distance between the points P and Q is : [2025]

1 $\sqrt{10}$

2 2

3 3

4 $2 \sqrt{3}$

Ans.
(3)

Vector parallel to line $L =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 2 \\ 1 & 3 & 4 \end{matrix} |$

$= 10 \hat{i} - 10 \hat{j} + 5 \hat{k} = 5 (2 \hat{i} - 2 \hat{j} + \hat{k})$

Equation of line L passing through point P(2, –1, 3) and parallel to vector $(2 \hat{i} - 2 \hat{j} + \hat{k})$ , is

$\frac{x - 2}{2} = \frac{y + 1}{- 2} = \frac{z - 3}{1} = λ$

$∴ x = 2 λ + 2, y = - 2 λ - 1, z = λ + 3$

$∵$ Line L intersects the yz-plane

$∴ 2 λ + 2 = 0 \Rightarrow λ = - 1$

Hence, point Q is (0, 1, 2)

Distance between point P(2, –1, 3) and Q(0, 1, 2)

$= \sqrt{{(0 - 2)}^{2} + {(1 + 1)}^{2} + {(2 - 3)}^{2}} = \sqrt{9} = 3$ .

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