Find the vector equation of a line passing through the point with position vector (2i -3j - 5k)

Question

Find the vector equation of a line passing through the point with position vector \((2\hat{i}-3\hat{j}-5\hat{k})\) perpendicular to the plane \(\vec{r}.(6\hat{i}-3\hat{j}+5\hat{k}) + 2=0.\) Also find the point of intersection of this line and the plane.

Answer 1

As the required line is perpendicular to the plane

So, the required line is parallel to \(\vec{n} = 6\hat{i}-3\hat{j}+5\hat{k}\)

Thus, the required line passes through the point with position vector \(\vec{a} = 2\hat{i}-3\hat{j}-5\hat{k}\) and is parallel to \(\vec{n} = 6\hat{i}-3\hat{j}+5\hat{k}.\)

Hence, the vector equation of the required line is \(\vec{r} = \vec{a} + \lambda\,\vec{n}\)

If the line (ii) meets the plane (i), then

Substituting \(\lambda = \frac{1}{35}\) in (ii), we get

Hence, the required point of intersection is \((\frac{76}{35},\frac{-108}{35},\frac{-170}{35})\) \(i.e.,\) \((\frac{76}{35},\frac{-108}{35},\frac{-34}{7}).\)