Question

Find the equation of the plane bisecting the line made by joing two points at right angle

Answer 1

he given points are A(2,3,4)andB(4,5,8)

The line segement AB is given by (x2−x1),(y2−y1),(z2−z1)

(ie)(4−2),(5−3),(8−4)

=(2,2,4)=(1,1,2)

Since the plane bisects AB at rightangles, AB−→− is the normal to the plane. which is n→

Therefore n→=i^+j^+2k^

Let C be the midpoint of AB.

(x1+x22,y1+y22,z1+z22)

=C(2+42,3+52,4+82)

=C(62,82,122)=(3,4,6)

Let this be a→=3i^+4j^+6k^

Hence the vector equation of the plane passing through C and ⊥ AB is

(r→−(3i^+4j^+6k^)).(i^+j^+2k^)=0

We know r→=xi^+yj^+zk^

=>[(xi^+yj^+zk^).(3i^+4j^+6k^)].(i^+j^+2k^)=0

On simplifying we get,

(x−3)i^+(y−4)j^+(z−6)k^).(i^+j^+2k^)=0

Applying the product rule we get,

(x−3)+(y−4)+2(z−6)=0

On simplifying we get,

x+y+2z=19

This is the required equation of the plane

THANKS!!