Question

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

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

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 $i + j - k\, i.e.,$ (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
$\overrightarrow{r} \cdot ( \hat{i} - \hat{j} + \hat{k} ) + 1 = 0$