JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 2)
The foot of perpendicular of the point (2, 0, 5) on the line $${{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}}$$ is ($$\alpha,\beta,\gamma$$). Then, which of the following is NOT correct?
$$\frac{\alpha}{\beta}=-8$$
$$\frac{\alpha \beta}{\gamma}=\frac{4}{15}$$
$$\frac{\beta}{\gamma}=-5$$
$$\frac{\gamma}{\alpha}=\frac{5}{8}$$
Explanation
$$ \mathrm{L}: \frac{\mathrm{x}+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text { (let) } $$
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$$ \begin{aligned} & \mathrm{P}(2 \lambda-1,5 \lambda+1,-\lambda-1) \\\\ & \overrightarrow{\mathrm{PA}}=(3-2 \lambda) \hat{\mathrm{i}}-(5 \lambda+1) \hat{\mathrm{j}}+(6+\lambda) \hat{\mathrm{k}} \\\\ & \text { Direction ratio of line } \Rightarrow \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\\\ & \text { Now, } \Rightarrow \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{b}}=0 \\\\ & \Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0 \\\\ & \Rightarrow \lambda=\frac{-1}{6} \\\\ & \mathrm{P}(2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv \mathrm{P}(\alpha, \beta, \gamma) \\\\ & \Rightarrow \alpha=2\left(-\frac{1}{6}\right)-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3} \\\\ & \Rightarrow \beta=5\left(-\frac{1}{6}\right)+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6} \\\\ & \Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6} \end{aligned} $$
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