Question

Find the equation of the plane which passes through the points [6,2,−4] and [3,−4,1] and parallel to the line joining the point [1,0,3] and [−1,2,4]

Answer 1

Answer:

Let the equation of plane passing through (−1,2,1) be

a(x+1)+b(y−2)+c(z−1)=0...(i)

Now the equation of line joining point (−3,1,2) and (2,3,4) is

2+3

x+3

=

3−1

y−1

=

4−2

z−2

⇒

5

x+3

=

2

y−1

=

2

z−2

...(ii)

∵ Plane (i) is perpendicular to line (ii)

⇒

5

a

=

2

b

=

2

c

=λ(say) [∵ Normal vector of (i) is parallel to line (ii) ]

⇒a=5λ,b=2λ,c=2λ

Hence, equation of required plane is

5λ(x+1)+2λ(y−2)+2λ(z−1)=0

⇒5x+5+2y−4+2z=0

⇒5x+2y+2z−1=0

If d is the distance from origin to plane then

d=

∣

∣

∣

∣

∣

5

2

+2

2

+2

2

0.x+0.y+0.z−1

∣

∣

∣

∣

∣

=

33

1