Question

Find the equation of the plane through the points (2,2,1)and (9,3,6) and perpendicular to the plane 2x+6y+6z=9

Answer 1

The equation of the plane passing through (2,2,1)(2,2,1)is a(x−2)+b(y−2)+c(z−1)=0a(x−2)+b(y−2)+c(z−1)=0-----(1)

Since this passes through (9,3,6)(9,3,6)

a(9−2)+b(3−2)+c(6−1)=0a(9−2)+b(3−2)+c(6−1)=0

(ie) 7a+b+5c=07a+b+5c=0---------(2)

Now the plane (1) is perpendicular to 2x+6y+6z=92x+6y+6z=9

∴∴2a+6b+4c=0$-------(3)

Solve equation (2) and (3) we get,

a∣∣∣1654∣∣∣a|1564|=b∣∣∣5472∣∣∣b|5742|=c∣∣∣7216∣∣∣c|7126|=λ=λ

(ie)a4−30=b10−28=c42−2a4−30=b10−28=c42−2=λ=λ

From equation (1) we get,

−26λ(x−2)−18λ(y−2)+40λ(z−1)=0−26λ(x−2)−18λ(y−2)+40λ(z−1)=0

−26x+52−18y+36+40z−40=0−26x+52−18y+36+40z−40=0

(ie) −26x−18y+40z+48=0−26x−18y+40z+48=0

Dividing throughout by −2−2 we get,

13x+9y−20z−24=013x+9y−20z−24=0

This is the required equation of the plane