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
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
C.
circle of radius
(2x-3y +4) +k (x-2y+3) = 0 is family of lines passing through (1,2). By congruency of triangles, we can prove that mirror image (h,k) and the point (2,3) will be equidistant from (1,2).
Therefore, Locus of (h,k) is PR = PQ
⇒ (h-1)2 + (k-2)2 = (2-1)2 + (3-2)2
(x-1)2 + (y-2)2 = 2
Locus is a circle of radius =