Question

Find the parametric and symmetric equation of the line passing through the point $(2,3,4)$ and perpendicular to the plane $5x+6y-7z=20$ ?

• A. $(x=5t+2,y=6t+3,z=-7t+4)$
  $\left(\frac{x-2}{5}=\frac{y-3}{6}=\frac{z-4}{7}\right)$
• B. $(x=5t+2,y=6t+3,z=-7t+4)$
  $\left(\frac{x-2}{5}=\frac{y-3}{-6}=\frac{x-4}{-7}\right)$
• C. $(x=5t+2,y=-6t+3,z=-7t+4)$
  $\left(\frac{x-2}{5}=\frac{y-3}{-6}=\frac{z-4}{-7}\right)$
• D. $(x=5t+2,y=6t+3,z=-7t+4)$
  $\left(\frac{x-2}{5}=\frac{y-3}{6}=\frac{z-4}{-7}\right)$

Answer 1

The correct option is D $(x=5t+2,y=6t+3,z=-7t+4)$

$\left(\frac{x-2}{5}=\frac{y-3}{6}=\frac{z-4}{-7}\right)$

The vector perpendicular to the plane, $5x+6y-7z=20$ is:

$\overrightarrow{n}=5\hat{i}+6\hat{j}-7\hat{j}$

This allows us to write the point-vector form of the line passing through the point $(2,3,4);$
$\overrightarrow{L}=2\hat{i}+3\hat{j}+4\hat{k}+t(5\hat{i}+6\hat{j}-7\hat{j})$
From the point-vector form we can extract the 3 parametric equations by observation:

$x=5t+2y=6t+3z=-7t+4$

To find the symmetric form we solve each of the parametric equations for t and then set them equal:

$t=\frac{x-2}{5}t=\frac{y-3}{6}t=\frac{z-4}{-7}$

Setting them equal gives us the symmetric form:

$\frac{x-2}{5}=\frac{y-3}{6}=\frac{z-4}{-7}$