Find the angle alpha in radians between the planes -4x + 4y - 4z =1 and 2x - 2y + 5z = 1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the normal vectors to the planes and then you need to find the angle between the normal vectors, such that:

`bar u = <-4,4,-4>` (vector normal to the plane `-4x + 4y - 4z = 1` )

`bar v = <2,-2,5> ` (vector normal to the plane `2x - 2y + 5z = 1` )

You need to evaluate the dot product to find the angle between the normal vectors, such that:

`bar u*bar v = |bar u|*|bar v|*cos alpha`

`alpha` - angle between the normal vectors bar u and bar v

`bar u*bar v = -4*2 + 4*(-2) + (-4)*5`

`bar u*bar v = -8 - 8 - 20 = -36`

`|bar u| = sqrt((-4)^2 + 4^2 + (-4)^2) => |bar u| = 4sqrt3`

`|bar v| = sqrt(2^2 + (-2)^2 + 5^2) => |bar v| = sqrt29`

`cos alpha = (bar u*bar v)/(|bar u|*|bar v|)`

`cos alpha = (-36)/(4sqrt(3*29)) => cos alpha = -9/sqrt(87) = -0.964 => alpha = cos^(-1)(-0.964)`

`alpha = (4pi)/5`

Hence, evaluating the angle between the planes, using the normal vectors to the planes, yields `alpha = (4pi)/5.`

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