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
If the angle between the line x = and the plane x + 2y + 3z = 4 is cos-1
then λ equal
2/3
3/2
2/5
2/5
Statement-1 : The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line:
Statement-2: The line: bisects the line segment joining A(1, 0, 7) and B(1, 6, 3).
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false.
Statement-1 is true, Statement-2 is false.
A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle θ with the positive z-axis, then θ
30°
45°
60°
60°
The line L given by passes through the point (13, 32). The line K is parallel to L and has the equation
. Then the distance
between L and K is
Let k be an integer such that triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point
The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point
C.
y'= 1 at point (0, 1)
∴ Slope of normal is –1
Hence equation of normal is x + y = 1
∴ (1/2, 1/2)satisfy it.