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
B.
10√3
the equation of the line passing through the point (1,5-9 and parallel to x =y=z is
Thus, any point on this line is of the form
(λ +1, λ-5 ,λ+9)
Now, if P (λ +1, λ-5, λ+9) is the point of intersection of line and plane, then
(λ+1) - (λ-5) +λ+9 = 5
λ +15 = 5
λ = -10
therefore coordinates of point P are (-9, -15,-1)
Hence, required distance
=
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