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
C.
circle of radius
(2x-3y +4) +k (x-2y+3) = 0 is family of lines passing through (1,2). By congruency of triangles, we can prove that mirror image (h,k) and the point (2,3) will be equidistant from (1,2).
Therefore, Locus of (h,k) is PR = PQ
⇒ (h-1)2 + (k-2)2 = (2-1)2 + (3-2)2
(x-1)2 + (y-2)2 = 2
Locus is a 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
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