Find the distance of the point (- 1, - 5,- 10) from the point of intersection of the line $r = 2^i -^j + 2^k + λ (3^i + 4^j + 2^k)$ and the plane $r . (^i -^j +^k) = 5$

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Solution

Given equation of line $r = 2^i -^j + 2^k + λ (3^i + 4^j + 2^k)$ . . . (i)
and the equation of plane $r . (^i -^j +^k) = 5$ . . . (ii)
Point of intersection of Eqs. (i) and (ii) is
$[2^i -^j + 2^k + λ (3^i + 4^j + 2^k)] . (^i -^j +^k) = 5 \Rightarrow 2 + 1 + 2 + λ (3 - 4 + 2) = 5 \Rightarrow 5 + λ = 5 \Rightarrow λ = 0$
On putting $λ$ = 0 in Eq. (i), we get $r = 2^i -^j + 2^k$
And r is the position vector of the point (2, -1, 2).
Distance between the point (-1, -5, -10) and (2, -1, 2)
$= \sqrt{(- 1 - 2)^{2} + (- 5 + 1)^{2} + (- 10 - 2)^{2}} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 u n i t s$

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(- 1, - 5, - 10)

\to r = 2^i -^j + 2^k + λ (3^i + 4^j + 2^k)

\to r \cdot (^i -^j +^k) = 5

(2, 12, 5)

\to r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)

\to r (^i - 2^j +^k) = 0

¯ r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)

¯ r \cdot (^i - 2^j +^k) = 0

\to r = 2^i - 4^j + 2^k + λ (3^i + 4^j + 2^k)

\to r . (^i - 2^j +^k) = 0

\to r = 2^i + 3^j -^k + λ (^i -^j + 2^k)

\to r = - 3^i + 4^j +^k + μ (^i -^j + 2^k)

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