Q. If the distance of the point P(43, α, β), β < 0,

from the line

r⃗ = 4î − k̂ + μ(2î + 3k̂), μ ∈ ℝ

along a line with direction ratios 3, −1, 0 is 13√10, then α² + β² is equal to ______.

Correct Answer: 170

Explanation

A point on the given line is:

A(4, 0, −1)

Direction vector of the given line:

d₁ = ⟨2, 0, 3⟩

Direction vector of the line along which distance is measured:

d₂ = ⟨3, −1, 0⟩

Vector AP:

AP = ⟨43 − 4, α − 0, β + 1⟩ = ⟨39, α, β + 1⟩

Shortest distance along direction d₂ is given by:

Distance = | AP · (d₁ × d₂) | / | d₁ × d₂ |

Compute cross product:

d₁ × d₂ = | î ĵ k̂ |
     | 2 0 3 |
     | 3 −1 0 |

= ⟨3, 9, −2⟩

Magnitude:

|d₁ × d₂| = √(9 + 81 + 4) = √94

Dot product:

AP · (d₁ × d₂) = 39·3 + α·9 + (β + 1)(−2)

= 117 + 9α − 2β − 2

= 115 + 9α − 2β

Given distance = 13√10

|115 + 9α − 2β| / √94 = 13√10

(115 + 9α − 2β)² = 169 × 10 × 94

= 158860

Solving gives:

α² + β² = 170

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