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
If a line makes an angle of π/4 with the positive directions of each of x-axis and y-axis, then the angle that the line makes with the positive direction of the z-axis is
π/6
π/3
π/4
π/4
If (2, 3, 5) is one end of a diameter of the sphere x2+ y2+ z2 − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are
(4, 9, –3)
(4, –3, 3)
(4, 3, 5)
(4, 3, 5)
Let A(h, k), B(1, 1) and C(2, 1) be the vertices of a right-angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which ‘k’ can take is given by
{1, 3}
{0, 2}
{–1, 3}
{–1, 3}
Let P = (−1, 0), Q = (0, 0) and R = ( 3, 3 √3) be three points. The equation of the bisector of the angle PQR