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
A.
We have,
⇒ 5k2 + 13k – 46 = 0
or
5k2 + 13k + 66 = 0 (no real solution exist)
∴ k =–23/5
or k = 2
k is an integer, so k =2
As
Therefore (2, 1/2)
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
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)