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Which one of the following planes is normal to the plane $3 x+y+z=5 ? \quad$
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Verified Answer
The correct answer is:
$x-2 y-z=6$
Direction cosines of the normal to the plane $3 x+y+z=5$ are $3,1,1$
Direction cosines of the normal to the plane $x-2 y-z=6$ are $1,-2,-1$
Sum of the product of direction cosines $=3 \times 1+1 \times(-2)+1 \times(-1)=0$
Hence, normals to the two planes are perpendicular to each other. Therefore two planes are alsoperpendicular.
Direction cosines of the normal to the plane $x-2 y-z=6$ are $1,-2,-1$
Sum of the product of direction cosines $=3 \times 1+1 \times(-2)+1 \times(-1)=0$
Hence, normals to the two planes are perpendicular to each other. Therefore two planes are alsoperpendicular.
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