Trigonometric EquationHard
Question
Two fixed points A(1, 2, 3) & B(7, 6, 5) are given. If P is a variable point on the plane x + y + z = 4 which is equidistant from points A & B, then locus of point P is -
Options
A.x + 2y + 3z = 24
B.x2 + y2 + z2 = 3
C.x - 10 = 
= z + 6
= z + 6D.
Solution
Plane perpendicular to AB & passing through mid point of AB i.e. M(4, 4, 4) is
6(x - 4) + 4(y - 4) + z(z - 4) = 0 3x + 2y + z = 24
required locus is line of intersection of planes
x + y + z = 4 & 3x + 2y + z = 24
i.e.x - 10 =
= z + 6
6(x - 4) + 4(y - 4) + z(z - 4) = 0 3x + 2y + z = 24
required locus is line of intersection of planes
x + y + z = 4 & 3x + 2y + z = 24
i.e.x - 10 =
= z + 6Create a free account to view solution
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