Find the equation of the plane which contains the line of intersection of the planes
and which is perpendicular to
the plane .
Find the equation of the line passing through the point (-1,3,-2) and
perpendicular to the lines
We know that, equation of a line passing through x1, y1, z1 with direction ratios a, b, c
So, the required equation of a line passing through ( - 1, 3, - 2 ) is:
Solving equation ( ii ) and ( iii ) by cross multiplication,
Putting the value of a, b and c in ( i ) gives
Find the equation of the plane determined by the point A( 3, - 1, 2 ), B( 5, 2, 4 ) and C( -1, -1, 6 ) and hence find the distance between the plane and the point P( 6, 5, 9 ).
ABC is formed by A(1, 8, 4), B (0, - 11, 4) and C(2, - 3,1). If D is the foot of the perpendicular from A to BC. Then the coordinates of D are
( - 4, 5, 2)
(4, 5, - 2)
(4, - 5, 2)
(4, - 5, - 2)
The distance of the point (1, −5, 9) from the plane x−y+z=5 measured along the line x=y=z is:
3√10
10√3
10/√3
20/3
Locus the image of the point (2,3) in the line (2x - 3y +4) + k (x-2y+3) = 0, k ε R is a
straight line parallel to X - axis
a straight line parallel to Y- axis
circle of radius
circle of radius