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Question
Solve the following :
Find the distance of the point (13, 13, – 13) from the plane 3x + 4y – 12z = 0.
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Solution
The distance of the point (x1, y1, z1) from the plane ax + by + cz + d = 0 is `|(ax_1 + by_1 + cz_1 + d)/sqrt(a^2 + b^2 + c^2)|`
∴ the distance of the point (1, 1, – 1) from the plane 3x + 4y – 12z = 0 is `|(3(1) + 4(1 - 12(-1)))/sqrt(3^2 + 4^2 + (-12)^2)|`
= `|(3 + 4 + 12)/sqrt(9 + 16 + 144)|`
= `(19)/sqrt(169)`
= `(19)/(13)`
= 19units.
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