ABC is formed by A(1, 8, 4), B (0, - 11, 4) and C(2, - 3,1). If D is the foot of the perpendicular from A to BC. Then the coordinates of D are
( - 4, 5, 2)
(4, 5, - 2)
(4, - 5, 2)
(4, - 5, - 2)
The distance of the point (1, −5, 9) from the plane x−y+z=5 measured along the line x=y=z is:
3√10
10√3
10/√3
20/3
Locus the image of the point (2,3) in the line (2x - 3y +4) + k (x-2y+3) = 0, k ε R is a
straight line parallel to X - axis
a straight line parallel to Y- axis
circle of radius
circle of radius
The distance of the point (1,0,2) from the point of intersection of the line and the plane x-y +z = 16 is
8
The equation of the plane containing the line 2x-5y +z = 3, x +y+4z = 5 and parallel to the plane x +3y +6z =1 is
2x + 6y + 12z = 13
x+3y+6z = -7
x+3y +6z = 7
x+3y +6z = 7
C.
x+3y +6z = 7
Let equation of plane containing the lines 2x- 5y +z = 3 and x+y+4z = 5 be
(2x-5y+z-3) + λ(x+y+4z-5) = 0
⇒ (2+λ)x + (λ-5)y + (4λ + 1)z -3 -5λ =0... (i)
This plane is parallel to the plane x +3y +6z = 1
On taking first two equalities, we get
6λ -30 = 3 + 12λ
-6λ = 33
λ = - 11/2
So, the equation of required plane is
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
3/2
5/2
7/2
7/2
If the lines
are coplanar, then k can have
any value
exactly one value
exactly two values
exactly two values