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Question
Solve the following :
Find the perpendicular distance of the origin from the plane 6x + 2y + 3z - 7 = 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 (0, 0, 0) from the plane 6x + 2y + 3z – 7 = 0 is
`|((6(0) + 2(0) + 3 (0) - 7))/sqrt(6^2 + 2^2 + 3^2)|`
= `|(0 + 0 + 0 - 7)/sqrt(36 + 4 + 9)|`
= `|(-7)/sqrt(49)|`
= `|(-7)/(7)|`
= 1units.
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