If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to line, x/1 = y/4= z/5 is Q, then PQ is equal to
The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal perpendicular to both the lines is
The lines p(p2+ 1) x – y + q = 0 and (p2+ 1)2x + (p2+ 1) y + 2q = 0 are
perpendicular to a common line for
no value of p
exactly one value of p
exactly two values of p
exactly two values of p
Let the line lie in the plane x + 3y – αz + β = 0. Then (α, β) equals
(6, – 17)
(–6, 7)
(5, –15)
(5, –15)
If P and Q are the points of intersection of the circles x2+ y2+ 3x + 7y + 2p – 5 = 0 and x2+ y2+ 2x + 2y – p2 = 0, then there is a circle passing through P, Q and (1, 1) for
all values of p
all except one value of p
all except two values of p
all except two values of p
The line passing through the points (5, 1, a) and (3, b, 1) crosses the yz−plane at the point . Then
a = 2, b = 8
a = 4, b = 6
a = 6, b = 4
a = 6, b = 4
Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle α with the positive x-axis, then cos α equals-
1/2
1
1
A.
If direction cosines of L be l, m, n, then
2l + 3m + n = 0
l + 3m + 2n = 0
The resultant of two forces P N and 3 N is a force of 7 N. If the direction of the 3 N force were reversed, the resultant would be N. The value of P is
5N
6N
4N
4N