Question

Show that the lines :

$\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (\hat{i}-\hat{j}+\hat{k})$

$\overrightarrow{r}=4\hat{j}+2\hat{k}+\mu (2\hat{i}-\hat{j}+3\hat{k})$ are coplanar.

Also, find the equation of the plane containing these lines.

Answer 1

$x\hat{i}+y\hat{j}+z\hat{k}=(1+\lambda )\hat{i}+(1-\lambda )\hat{j}+(1+\lambda )\hat{k}$

$x\hat{i}+y\hat{j}+z\hat{k}=(0+2\mu )\hat{i}+(4-\mu )\hat{j}+(2+3\mu )\hat{k}$

$\Rightarrow 1+\lambda =2\mu ...\left(i\right)$

$1-\lambda =4-\mu ...\left(ii\right)$

and $1+\lambda =2+3\mu ...\left(iii\right)$

$\Rightarrow \lambda =2\mu -1$

Put the value of$\lambda$ in eq. (ii), we have

$1-2\mu +1=4-\mu$

$-\mu =2$

$\mu =-2$

or $\lambda =2(-2)-1$

= $-4-1$

= $-5$

Substitute the value of $\lambda$ and $\mu$ in eq. (iii), we have

$1-5=2+3(-2)$

$-4=2-6$

$-4=-4$ which is true

Hence, they are coplanar.

Point of intersection is $(-4,6,-4)$

Equation of plane is

$a(x+4)+b(y-6)+c(z+4)=0...\left(i\right)$

$a-b+c=0...\left(ii\right)$

$2a-b+3c=0...\left(iii\right)$

From (ii) and (iii), we have

$\frac{a}{-3+1}=\frac{b}{2-3}=\frac{c}{-1+2}$

$\frac{a}{-2}=\frac{b}{-1}=\frac{c}{1}$

or $\frac{a}{2}=\frac{b}{1}=\frac{c}{-1}$

From (i), we have

$2(x+4)+y-6-(z+4)=0$

$\Rightarrow 2x+8+y-6-z-4=0$

$\Rightarrow 2x+y-z-2=0$