Find the equation of the line passing through a point $B$, with position vector $\mathbf{b}$ relative to the origin $O$, which is perpendicular to and intersects the line $\mathbf{r} = \mathbf{a} + \lambda \mathbf{c}$ where $\mathbf{c} \ne \mathbf{0}$, given that $B$ is not a point of the line.
No idea how to approach question. Is it that the dot product of line passing through $B$ and $\mathbf{r} =\mathbf{0}$?
If $B$ is not on $\mathbf{r}=\mathbf{a}+\lambda \, \mathbf{c}$, then $\mathbf{b} \ne \mathbf{a}$.
The point of intersection is given by
\begin{align*} 0 &= (\mathbf{r}-\mathbf{b}) \cdot \mathbf{c} \\ 0 &= \mathbf{a} \cdot \mathbf{c}+\lambda \, c^2-\mathbf{b} \cdot \mathbf{c} \\ \lambda &= \frac{(\mathbf{b}-\mathbf{a}) \cdot \mathbf{c}}{c^2} \end{align*}
So the equation of the required line is
$$\mathbf{r}= \mathbf{b}+\mu \left[ \mathbf{a}-\mathbf{b}+ \frac{(\mathbf{b}-\mathbf{a}) \cdot \mathbf{c}}{c^2} \, \mathbf{c} \right]$$
or equivalently in case of $\mathbb{R}^3$,
$$\mathbf{r}= \mathbf{b}+\frac{\mu \, \mathbf{c} \times [(\mathbf{a}-\mathbf{b}) \times \mathbf{c}]}{c^2}$$