Question

Find the vector equation of the straight line passing through (1, 2, 3) and perpendicular to the plane $r.(\hat{i}+2\hat{j}-5\hat{k})+9=0$

Answer 1

Given plane is $r.(\hat{i}+2\hat{j}-5\hat{k})+9=0$
In Cartesian form, this plane can be written as x + 2y - 5z + 9 =0.
Putting $r=x\hat{i}+y\hat{j}+z\hat{k}$
Direction ratio of any line perpendicular to the given plane are 1, 2, -5. If this line passes through (1, 2, 3), then its equation is
is $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{-5}(\because \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c})$
The above equation can be written in vector form
$r=(1\hat{i}+2\hat{j}+3\hat{k})+\lambda (1\hat{i}+2\hat{j}-5\hat{k})$
where, $\lambda$ being any real number.