A) \[x+2y+z=6\]
B) \[x-2y+z=6\]
C) \[x+2y-z=6\]
D) \[x-2y-z=6\]
Correct Answer: D
Solution :
[d] Direction cosines of the normal to the plane \[3x+y+z=5\]are 3, 1, 1 Direction cosines of the normal to the plane \[x-2y-z=6\] are 1, -2, -1 Sum of the product of direction cosines \[=3\times 1+1\times (-2)+1\times (-1)=0\] Hence, normal to the two planes are perpendicular to each other. Therefore two planes are also perpendicular.
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