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Question
Find the equation of the plane passing through the point (1, 1, 1) and is perpendicular to the line `("x" - 1)/3 = ("y" - 2)/0 = ("z" - 3)/4`. Also, find the distance of this plane from the origin.
Solution
Since the plane is perpendicular to the given line, its direction ratios are proportional to 3, 0, 4.
So, the required equation of the plane is
3x + 0y + 4z + d = 0 ......(i)
where d is a constant.
Since this plane passes through (1, 1, 1)
3 + 0 + 4 + d = 0
⇒ d = − 7
From equation (i), we get
3x + 4z − 7 = 0 .........(ii)
This is the required equation of the plane.
Perpendicular distance of (ii) from the origin is given by,
d = `|"ax"_1 + "by"_1 + "cz"_1 + "d"|/sqrt("a"^2 + "b"^2 + "c"^2)`
⇒ d = `|3 xx 0 + 4 xx 0 - 7|/sqrt(3^2 + 0^2 + 4^2)`
⇒ d = `|-7|/sqrt(9 + 16)`
⇒ d = `7/sqrt25`
= `7/5` units
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