Find the vector equation of the line passing through the point with position vector (i - 2j + 5k) and perpendicular to the plane

Question

Find the vector equation of the line passing through the point with position vector (\(\hat{i}\) - 2\(\hat{j}\) + 5\(\hat{k}\)) and perpendicular to the plane \(\bar{r}\). (2\(\hat{i}\) - 3\(\hat{j}\) - \(\hat{k}\)) = 0.

Answer 1

Given \(\bar{r}\)(2\(\hat{i}\) - 3\(\hat{j}\) - \(\hat{k}\)) = 0 and the vector has position vector (\(\hat{i}\) - 2\(\hat{j}\) + 5\(\hat{k}\))

To find – The vector equation of the line passing through (1, - 2, 5) and perpendicular to the given plane 

Tip – The equation of a plane can be given by \(\bar{r}\).\(\bar{n}\) = d where \(\bar{n}\) is the normal of the plane 

A line parallel to the given plane will be in the direction of the normal and will have the direction ratios same as that of the normal. Formula to be used – If a line passes through the point (a, b, c) and has the direction ratios as (a’, b’, c’), then its vector equation is given by \(\bar{r}\) = (a\(\hat{i}\) + b\(\hat{j}\) + c\(\hat{k}\)) + λ(a'\(\hat{i}\) + b'\(\hat{j}\) + c'\(\hat{k}\)) where λ is any scalar constant 

The required equation will be \(\bar{r}\) = (\(\hat{i}\) - 2\(\hat{j}\) + 5\(\hat{k}\)) + λ(2\(\hat{i}\) - 3\(\hat{j}\) - \(\hat{k}\)) for some scalar λ