If the distance of the point P (43, α, β), β<0 , from the line overrightarrow r =4 hat i -hat k +μ(2 hat i +3 hat k ), μ in R along a line with direction ratios 3,-1,0 is 13 √(10) , then α 2 +β 2 is equal to \ \ \ \ .

Mathematics · 3D Geometry

JEE Main 2026 — 28 January, Evening Shift — Question 24

If the distance of the point $\mathrm{P}(43, \alpha, \beta), \beta<0$ , from the line $\overrightarrow{\mathrm{r}}=4 \hat{\mathrm{i}}-\hat{\mathrm{k}}+\mu(2 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}), \mu \in \mathbf{R}$ along a line with direction ratios $3,-1,0$ is $13 \sqrt{10}$ , then $\alpha^{2}+\beta^{2}$ is equal to $\_\_\_\_$ .

Answer: 170

Numerical answer — enter this value.

Step-by-step solution

$\frac{x-43}{3}=\frac{y-\alpha}{-1}=\frac{z-\beta}{0} \Rightarrow P_{1}(43+3 \lambda, \alpha-\lambda, \beta)$

$\frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3} \Rightarrow P_{1}(2 \mu+4,0,3 \mu-1)$

$\therefore \mu=\frac{3 \lambda+39}{2}, \alpha=\lambda, \beta=\frac{9 \lambda-115}{2}$

$\mathrm{P}(43, \alpha, \beta), \mathrm{P}_{1}(43+3 \alpha, 0, \beta)$

$\left(\mathrm{PP}_{1}\right)^{2}=1690=10 \alpha^{2},$ $\therefore \alpha=13, \beta=1$

$\therefore \alpha^{2}+\beta^{2}=170$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Direction Cosines and Direction Ratios