Equation of the line through the point (2, 3, 1) and parallel to the line of intersection of the planes x - 2y - z + 5 = 0 and x + y + 3z = 6 is
Foot of the perpendicular drawn from the origin to the plane 2x - 3y + 4z = 29 is
(5, - 1, 4)
(7, - 1, 3)
(5, - 2, 3)
(2, - 3, 4)
The projection of the line segment joining (2, 0, - 3) and (5, - 1, 2) on a straight line whose direction ratios are 2, 4, 4, is
The equation of the plane which bisects the· line segment joining the points (3, 2, 6) and (5, 4, 8) and is perpendicular to the same line segment, is
x + y + z = 16
x + y + z = 10
x + y + z = 12
x + y + z = 14
D.
x + y + z = 14
Since, plane passes through mid-point of (3, 2, 6) and (5, 4, 8).
Hence, will lie on the plane. Also, plane is perpendicular to the line segment joining (3, 2, 6) and (5, 4, 8). Thus, DR's of the normal will be 5 - 3, 4 - 2, 8 - 6 i.e. 2 : 2 : 2 or 1 : 1 : 1. Hence, required equation of the plane will be
1(x - 4) + 1(y - 3) + 1(z - 7) = 0
x + y + z = 14
The foot of the perpendicular from the point (1, 6, 3) to the line is
(1, 3, 5)
(- 1, - 1, - 1)
(2, 5, 8)
(- 2, - 3, - 4)