Question

The line of intersection of the planes $\overrightarrow{r}\cdot (3\hat{i}-\hat{j}+\hat{k})=1$ and $\overrightarrow{r}\cdot (\hat{i}+4\hat{j}-2\hat{k})=2$, is.

• A. $\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{-7}=\frac{z}{-13}$
• B. $\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{7}=\frac{z}{-13}$
• C. $\frac{x-\frac{4}{7}}{-2}=\frac{y}{7}=\frac{z-\frac{5}{7}}{13}$
• D. $\frac{x-\frac{4}{7}}{2}=\frac{y}{-7}=\frac{z+\frac{5}{7}}{13}$

Answer 1

The correct option is A $\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{-7}=\frac{z}{-13}$

$\overrightarrow{n}=\overrightarrow{n_1}\times \overrightarrow{n_2}$

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ 1& 4& -2\end{array}\right|$

$=\hat{i}(-2)-\hat{j}(-7)+\hat{k}\left(13\right)$

$\overrightarrow{n}=2\hat{i}+7\hat{j}+13\hat{k}$

Now
$3x-y+z=1$
$x+4y-2z=1$

but $z=0$

$3x-y=1\times 4$

$x+4y=2$

$13x=6$

$x=\frac{6}{13}$

$y=\frac{5}{13}$
is

$\frac{x-\frac{6}{13}}{-2}=\frac{y-\frac{5}{13}}{7}=\frac{z-0}{13}$

or

$\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{-7}=\frac{z}{-13}$