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
The angle between a normal to the plane 2x - y + 2z - 1 = 0 and the Z-axis is
C.
Given equation of plane is
Zx - y + 2z - 1 = 0 ...(i)
DR's of normal to the given plane is (2, - 1, 2)
Given plane meet the z-axis.
Put x = 0, y = 0 in Eq. (i), we get
2(0) - 0 + 2z - 1 = 0
So, the point on z-axis is .
Angle between normal to the plane and z-axis,
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
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)