JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let two vertices of a triangle $ABC$ be $(2, 4, 6)$ and $(0, -2, -5)$ , and its centroid be $(2, 1, -1)$ . If the image of the third vertex in the plane $x + 2y + 4z = 11$ is $(\alpha, \beta, \gamma)$ , then $\alpha\beta + \beta\gamma + \gamma\alpha$ is equal to:

Choose the correct answer:

Explanation

Find the third vertex $C(x_3, y_3, z_3)$ :

Using the centroid formula $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$ :
- $2 = \frac{2 + 0 + x_3}{3} \implies x_3 = 4$
- $1 = \frac{4 - 2 + y_3}{3} \implies y_3 = 1$
- $-1 = \frac{6 - 5 + z_3}{3} \implies z_3 = -4$
  
  So, $C = (4, 1, -4)$ .
Find the image $(\alpha, \beta, \gamma)$ of $(4, 1, -4)$ in $x + 2y + 4z = 11$ :

Using the image formula $\frac{x-\alpha}{a} = \frac{y-\beta}{b} = \frac{z-\gamma}{c} = -2 \frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$ :

$\frac{\alpha-4}{1} = \frac{\beta-1}{2} = \frac{\gamma+4}{4} = -2 \frac{(4) + 2(1) + 4(-4) - 11}{1^2+2^2+4^2}$

$\frac{\alpha-4}{1} = \frac{\beta-1}{2} = \frac{\gamma+4}{4} = -2 \frac{-21}{21} = 2$
- $\alpha = 4 + 2 = 6$
- $\beta = 1 + 4 = 5$
- $\gamma = -4 + 8 = 4$
Calculate $\alpha\beta + \beta\gamma + \gamma\alpha$ :

$(6 \times 5) + (5 \times 4) + (4 \times 6) = 30 + 20 + 24 = 74$

Choose the correct answer:

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