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)
The plane x + 3y + 13 = 0 passes through the line of intersection of the planes 2x - 8y + 4z = p and 3x - 5y + 4z + 10 = 0. If the plane is perpendicular to the plane 3x - y - 2z - 4 = 0, then the value of p is
2
5
9
3
If a straight line makes the angles 60°, 45° and with X, Y and Z-axes respectively, then is equal to
1
The direction cosines of the straight line given by the planes x = 0 and z = 0 are
1, 0, 0
0, 0, 1
1, 1, 0
0, 1, 0