Question

Find the equation of the plane through the points (3,2,2) , (1,0,-1), and parallel to the line $\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-2}{3}$. Convert to vector form.

Answer 1

Equation of a plane passing through ($3,2,2$) is

$A(x-3)+B(y-2)+C(z-2)=0$ (1)

Since it is passing through ($1,0,-1$)

So, $A(1-3)+B(0-2)+C(-1-2)=0$

$-2A+2B+3C=0$ (2)

Since plane (1) is parallel to the line

$\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-2}{3}$

so, $1\times A-2\times B+3\times C=0$

$A-2B+3C=0$ (3)

Adding (2)and (3)

$3A+6C=0$

$3A=-6C$

$A=-2C$

Subtracting (3)from (2)

$A+4B=0$

$-2C+4B=0$ ($A=-2C$)

$4B=2C$

$B=\frac{1}{2}C$

Putting values of A and B in (1)

$-2c(x-3)+\frac{1}{2}c(y-2)+c(z-2)=0$

$4c(x-3)-c(y-2)-2c(z-2)=0$

$4(x-3)-(y-2)-2(z-2)=0$

$4x-12-y+2-2z+4=0$

$4x-y-2z=6$

$(x\hat{i}+y\hat{j}+z\hat{k})\times (4\hat{i}-\hat{j}-2\hat{k})=6$

$\overrightarrow{r}\times (4\hat{i}-\hat{j}-2\hat{k})=6$