The cartesian equation of the line is . The vector equation of the line is
-1/√6
1/√6
2/√6
-2/√6
The value of p and q so that the points (p, q,1), (-1, 4, -2) and (0, 2, -1) are collinear are
-2, -2
2, 2
2, -2,
0, 2
The value of λ for which the point A(2, 1, 3), B(5, 0, 5) and C(-4, λ, -1) are collinear is
1
2
3
4
Equation of the line passing through the point (2, 3, 4) and (4, 6, 5) is
Find the vector equation for the line passing through the point (-1, 0, 2) and (3, 4, 6) is
The equation of a line passing through the point(-1, 2, 3) and having direction ratios proportional to -4, 5, 6 is
None of these
Find the vector equation of the line joining the points whose position vectors are
The co ordinates of the point where the line through A(5, 1, 6) and B (3, 4, 1) cross the yz - plane
(17/2, 0, -13/2)
(0, 0, -13/2)
(17/2, 0, 1)
(0, 17/2, -13/2)