Question

Equation of the plane containing two intersecting lines $\overrightarrow{r}=\hat{i}+\hat{j}-\hat{k}+\lambda (2\hat{i}+\hat{j}-\hat{k})$ and$\overrightarrow{r}=\hat{i}+\hat{j}-\hat{k}+\mu (\hat{i}+2\hat{j}+\hat{k})$

• A. $\overrightarrow{r}\cdot \left(\hat{i}-\hat{j}+\hat{k}\right)+1=0$
• B. $\overrightarrow{r}\cdot \left(\hat{i}-\hat{j}+\hat{k}\right)=1$
• C. $x-y+z+1=0$
• D. $\overrightarrow{r}\cdot \left(\hat{i}+\hat{j}-\hat{k}\right)=0$

Answer 1

Answer: (C)

Since the plane contains the given lines, so it also contains the point of intersection $\hat{i}+\hat{j}-\hat{k}$.
Let the plane be $ax+by+cz+d$ = 0
Since it contains the given lines
$\therefore$ $2a+b-c$ = 0 and $a+2b+c$ = 0
$\Rightarrow$ $\frac{a}{1+2}=\frac{b}{-1-2}=\frac{c}{4-1}$
$\therefore$ Plane is $3x-3y+3z+d$ = 0
Since it passes through $i+j-ki.e.,$ (1, 1, -1)
$\therefore$ 3 - 3 - 3 + $d$ = 0 $\Rightarrow d$ = 3
$\therefore$ plane is $3x-3y+3z+3$ = 0
or $x-y+z+1$ = 0 which can also be written as
$\overrightarrow{r}\cdot (\hat{i}-\hat{j}+\hat{k})+1=0$