Three Dimensional Geometry
Perpendiculars are drawn from points on the line 2x+2=−1y+1=3z to the plane x+y+z=3 The feet of perpendiculars lie on the line (a) 5x=8y−1=−13z−2 (b) 2x=3y−1=−5z−2 (c) 4x=3y−1=−7z−2 (d) 2x=−7y−1=5z−2
Let A(a⃗ ) and B(b⃗ ) be points on two skew line r⃗ =a⃗ +λ⃗ and r⃗ =b⃗ +uq⃗ and the shortest distance between the skew line is 1, where p⃗ and q⃗ are unit vectors forming adjacent sides of a parallelogram enclosing an area of 12units. If an angle between AB and the line of shortest distance is 60∘, then AB=
The locus of a point, such that the sum of the squares of its distances from the planes x+y+z=0,x−z=0 And x−2y+z=0is 9, is
The vector equation of the line of intersection of the planes r⃗ =b⃗ +λ1(b⃗ −a⃗ )+μ1(a⃗ −c⃗ ) and r⃗ =b⃗ +λ2(b⃗ −c⃗ )+μ2(a⃗ +c⃗ )a⃗ ,b⃗ ,c⃗ being non-coplanar vectors, is
Find the vector equation of line passing through the point (1,2,−4) and perpendicular to the two lines: 3x−8=−16y+19=7z−10and3x−15=8y−29=−5z−5
Find the vector equation of the line passing through (1,2,3) and parallel to the planes ri^−j^+2k^˙andr3i^+j^+k^˙=6.