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Question
Find the distance of the point (2, 3, 4) measured along the line `(x - 4)/3 = (y + 5)/6 = (z + 1)/2` from the plane 3x + 2y + 2z + 5 = 0.
Solution
Here, Equation of line PQ is `(x - 2)/3 = (y - 3)/6 = (z - 4)/2` = λ(let)
∴ Q(3λ + 2, 6λ + 3, 2λ + 4)
Point Q lies on the plane 3x + 2y + 2z + 5 = 0
∴ 3(3λ + 2) + 2(6λ + 3) + 2(2λ + 4) + 5 = 0
⇒ 9λ + 6 + 12λ + 6 + 4λ + 8 + 5 = 0
⇒ 25λ = –25
⇒ λ = –1
So, Q(–1, –3, 2)
Now, distance between points P(2, 3, 4) and Q(–1, –3, 2) is given by
d = `sqrt((1 - 1 - 2)^2 + (-3 - 3)^2 + (2 - 4)^2)`
= `sqrt(9 + 36 + 4)`
= `sqrt(49)`
= 7 units
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