Q47 / 75 · JEE Main 2025 (4 Apr Shift 1) · Mathematics
Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is
- A42
- Correct answerB46
- C48
- D40
Solution
$\begin{aligned} & {\mathrm{A}(3, \alpha, 3) \& B(-3,-7, \beta)} \\ & \overrightarrow{\mathrm{BA}}=6 \hat{\mathrm{i}}+(\alpha+7) \hat{\mathrm{j}}+(3-\beta) \hat{\mathrm{k}} \\ \end{aligned}$
$\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|$
$\frac{|\overrightarrow{\mathrm{BA}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}=3 \sqrt{30}$
$36+15(\alpha+7)-3(3-\beta)=(3 \sqrt{30})^2$
$\begin{aligned} & 36+15 \alpha+105-9+3 \beta=270 \\ & 15 \alpha+3 \beta=138 \\ & 5 \alpha+\beta=46\end{aligned}$
$\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|$
$\frac{|\overrightarrow{\mathrm{BA}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}=3 \sqrt{30}$
$36+15(\alpha+7)-3(3-\beta)=(3 \sqrt{30})^2$
$\begin{aligned} & 36+15 \alpha+105-9+3 \beta=270 \\ & 15 \alpha+3 \beta=138 \\ & 5 \alpha+\beta=46\end{aligned}$
Hence, the correct answer is option: 'B'.