VectorHard
Question
A plane P passes through a point Q(3, - 2, 1) and is perpendicular to the vector 
. The distance between the plane P and the plane 
+ 33 = 0, equals
. The distance between the plane P and the plane + 33 = 0, equalsOptions
A.3
B.2
C.1
D.
Solution
Equation of the plane passing through the point Q(3, - 2, 1) is
A(x - 3) + B(y + 2) + C(z - 1) = 0 ....(1)
equation (1) is perpendicular to
henceA = 4; B = 7 and C = - 4
hence equation of the plane is
4x + 7y – 4z + 6 = 0 ....(2)
given 4x + 7y - 4z + 33 = 0 ....(3)
∴ Distance between (2) and (3)
=
= 0
A(x - 3) + B(y + 2) + C(z - 1) = 0 ....(1)
equation (1) is perpendicular to
henceA = 4; B = 7 and C = - 4
hence equation of the plane is
4x + 7y – 4z + 6 = 0 ....(2)
given 4x + 7y - 4z + 33 = 0 ....(3)
∴ Distance between (2) and (3)
=
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