1.

Find the distance of the plane `3x- 4y+12 z=3`from theorigin.

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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