Q. 275.0( 3 Votes )

Find the distance

Answer :

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 =


=


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

Find the equationMathematics - Board Papers

Find the equationMathematics - Board Papers

Find the length oMathematics - Board Papers

Find the coordinaMathematics - Board Papers

Find the equationMathematics - Board Papers

Show that the linMathematics - Board Papers