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
If the line lies in the plane lx +my -z = 9, then l2 +m2 is equal to
26
18
5
2
D.
2
Since the line lies in the plane lx my-z =9, therefore we have 2l-m-3 = 0
[∴ normal will be perpendicular to the line]
⇒ 2l-m = 3
and 3l -2m +4 = 9
[∴ point (-3,-2-4) lies on the plane
⇒ 3l-2m = 5
On solving eqs (i) adn (ii), we get
l =1 and m=-1
therefore, l2 +m2 = 2
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