An angle between the plane x+y+z=5 and the line of intersection of the planes 3x+4y+z−1=0 and 5x+8y+2z+14=0, is
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Official solution
Normal to 3x+4y+z=1 is \[3\widehat i + 4\widehat j + \widehat k\]
Normal to 5x+8y+2z=−14 is \[5\widehat i + 8\widehat j + 2\widehat k\]
The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors.
Therefore the direction ratios of the line is \[ - \widehat j + 4\widehat k\]
Hence the angle between the plane x + y + z + 5 = 0
and the intersection line is \[{\sin ^{ - 1}}\left( {{{ - 1 + 4} \over {\sqrt {17} \sqrt 3 }}} \right) = {\sin ^{ - 1}}\left( {\sqrt {{3 \over {17}}} } \right)\]