Q27 of 45 Page 1

Find the distance between the point (7, 2, 4) and the plane determine by the points A(2, 5, –3),B(–2, –3, 5) and C(5, 3, - 3).

OR


Find the distance of the point (-1, - 5, - 10) from the point of intersection of the lineand the plane

The equation of plane through points A(2, 5, –3),B(–2, –3, 5) and C(5, 3, - 3) is


=0


⇒=0


⇒ (x-2){0+16}-(y-5){0-24}+(z+3){8+24} = 0


⇒ 16x-32+24y-120+32z+96=0


⇒ 16x+24y+32z-56=0


⇒ 2x+3y+4z-7=0


Now, distance of plane from (7,2,4) =


=


OR


Given line,


And plane,


Intersection point of line and plane can be obtained by solving both equations.


Now, substituting the value of in equation of plane.


⇒


⇒ (2+1+2) +λ (3-4+2) = 5


⇒ 5+λ =5


⇒ λ =0


Thus, the position vector of intersecting point is and point of intersection is (2,-1,2).


The required distance =


=


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