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)
D.
(2, - 3, 4)
Let the foot of the perpendicular in the 2x - 3y + 4z = 29 be P.
So, the point satisfy the given plane.
Since, OF is perpendicular to the given plane. Therefore, normal to the plane is parallel to OF.
On putting the value of in Eq.(i), we get
2(2k) - 3(- 3k) + 4(4k) = 29
Hence, foot of perpendicular is (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)