JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The plane $2x - y + z = 4$ intersects the line segment joining $A(a, -2, 4)$ and $B(2, b, -3)$ at the point $C$ in the ratio $2:1$ and the distance of $C$ from the origin is $\sqrt{5}$ . If $ab < 0$ and $P$ is the point $(a - b, b, 2b - a)$ then $CP^2$ is equal to:

Choose the correct answer:

Explanation

Solution

Point $C$ section formula se: $C = \left(\frac{2(2)+1(a)}{3}, \frac{2(b)+1(-2)}{3}, \frac{2(-3)+1(4)}{3}\right) = \left(\frac{4+a}{3}, \frac{2b-2}{3}, \frac{-2}{3}\right)$ .

$C$ plane par hai: $2(\frac{4+a}{3}) - (\frac{2b-2}{3}) + (\frac{-2}{3}) = 4 \implies 8 + 2a - 2b + 2 - 2 = 12 \implies 2a - 2b = 4 \implies a - b = 2$ .

Distance $OC = \sqrt{5} \implies \frac{(a+4)^2}{9} + \frac{(2b-2)^2}{9} + \frac{4}{9} = 5$ .

Solve karne par $a=3, b=1$ (fails $ab<0$ ) or $a=1, b=-1$ (satisfies $ab<0$ ).

$C = (\frac{5}{3}, \frac{-4}{3}, \frac{-2}{3})$ , $P = (2, -1, -3)$ .

CP^2 = (2-\frac{5}{3})^2 + (-1+\frac{4}{3})^2 + (-3+\frac{2}{3})^2 = \frac{1}{9} + \frac{1}{9} + \frac{49}{9} = \frac{51}{9} = \frac{17}{3}

Sahi Option: (2)

Choose the correct answer:

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