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

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Solution

The plane can be written as $x - y + z = 5$ . Any point on the line is $(2 + 3 λ, 4 λ - 1, 2 + 2 λ)$ .
Now, for point of intersection, substitute equation of point in plane to obtain parameter $λ$ .
$∴ 2 + 3 λ - (4 λ - 1) + 2 + 2 λ = 5$
$∴ 5 + λ = 5$
$∴ λ = 0$
Hence, point of intersection of line and plane is $(2, - 1, 2)$ .
Distance between $(- 1, - 5, - 10)$ and $(2, - 1, 2) = \sqrt{(2 + 1)^{2} + (- 1 + 5)^{2} + (2 + 10)^{2}} = 13$
This is the required answer.

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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.

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