Question

The equations of the line of intersection of the planes $x+y+z=2$ and $3x-y+2z=5$ in symmetric form are

• A. $\frac{x-\frac{7}{4}}{4}=\frac{y-\frac{1}{4}}{-1}=\frac{z}{-3}$
• B. $\frac{x}{3}=\frac{y+\frac{1}{3}}{1}=\frac{z-\frac{7}{4}}{-4}$
• C. $\frac{x}{1}=\frac{3y+1}{1}=\frac{3z-7}{-4}$
• D. none of these

Answer 1

Answer: (B)

$\frac{x}{3}=\frac{y+\frac{1}{3}}{1}=\frac{z-\frac{7}{4}}{-4}$

Given, $\overrightarrow{n_1}=\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{n_2}=3\hat{i}-\hat{j}+2\hat{k}$

Therefore the direction vector of the line is parallel to

$\overrightarrow{n_1}\times \overrightarrow{n_2}$
$=(\hat{i}+\hat{j}+\hat{k})\times (3\hat{i}-\hat{j}+2\hat{k})$
$=3\hat{i}+\hat{j}-4\hat{k}$

Hence, the equation of the line will be of the form

$\frac{x-\alpha }{3}=\frac{y-\beta }{1}=\frac{z-\gamma }{-4}$