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Find the distance of the plane `3x- 4y+12 z=3`from theorigin. |
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Answer» Distance of a plane from origin can be given as, `D = |d/sqrt(a^2+b^2+c^2)|` In the given equation, `a = 3, b = -4, c = 12, d =-3` `D =|(-3)/sqrt(3^2+(-4)^2+12^2)| = |(-3)/sqrt(9+16+144)| = |(-3)/sqrt169|` `=>D = 3/13` So, distance of the given plane from the origin is `3/13`. |
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