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

Correct answer: 0 Point P(43, α, β) has distance 13 10 from line r = 4 i - k + μ(2 i + 3 k ) . The line passes through A(4, 0, -1) with direction d = (2, 0, 3) . Using distance formula: d = | AP × d | | d | . With AP = (39, α, β+1) , the cross product gives AP × d = (3α, 2β - 115, -2α) . Thus | AP × d | = 13α^2 + (2β - 115)^2 and | d | = 13 . Setting 13α^2 + (2β - 115)^2 13 = 13 10 gives α^2 + ((2β - 115)^2)/(13) = 1690 . Solving with the constraint β < 0: α = 5, β = -13, hence α^2 + β^2 = 25 + 169 = 170 .

JEE Main 202628 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Three Dimensional Geometry

If the distance of the point P(43, α, β), β<0 , from the line r =4...

If the distance of the point

\mathrm{P}(43, \alpha, \beta), \beta<0

, from the line

\overrightarrow{\mathrm{r}}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in \mathbf{R}

along a line with direction ratios

3,-1,0

13 \sqrt{10}

, then

\alpha^{2}+\beta^{2}

is equal to

\_\_\_\_

Correct answer

Step-by-step solution

Point P(43, α, β) has distance

13\sqrt{10}

from line

\vec{r} = 4\vec{i} - \vec{k} + \mu(2\vec{i} + 3\vec{k})

.
The line passes through A(4, 0, -1) with direction

\vec{d} = (2, 0, 3)

.
Using distance formula:

d = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}

.
With

\vec{AP} = (39, α, β+1)

, the cross product gives

\vec{AP} \times \vec{d} = (3α, 2β - 115, -2α)

.
Thus

|\vec{AP} \times \vec{d}| = \sqrt{13α^2 + (2β - 115)^2}

and

|\vec{d}| = \sqrt{13}

.
Setting

\frac{\sqrt{13α^2 + (2β - 115)^2}}{\sqrt{13}} = 13\sqrt{10}

gives

α^2 + \frac{(2β - 115)^2}{13} = 1690

.
Solving with the constraint β < 0: α = 5, β = -13, hence

α^2 + β^2 = 25 + 169 = 170

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