Use the dot/scalar product to solve the problem
Line 1 has vector equation $(2\mathrm{i}-\mathrm{j}) + \lambda(3\mathrm{i} + 2\mathrm{j})$
Find the vector equation of the line perpendicular to Line 1 and passing through the point with position vector $(4\mathrm{i} + 3\mathrm{j})$.
I can solve this problem by converting Line 1 into cartesian equation, but I dont know how to use the dot/scalar product to solve it.
Best Answer
Keeping it abstract, you have a line $L_1$ given by $\vec P + \lambda \vec v$, and you want to find a line $L_2$ perpendicular to $L_1$ and passing through an external point $\vec E$.
We want to find the point $\vec C$ on $L_1$ closest to $\vec E$, because then $L_2$ will be the line through $\vec C$ and $\vec E$. The vector $\vec{CE}$ will be perpendicular to $L_1$, i.e. perpendicular to $\vec v$.
Finding $\lambda_C$ for $\vec C$: $$\vec{CE}\cdot\vec v= \vec 0\\ (\vec E-(\vec P+\lambda_C \vec v))\cdot \vec v = \vec 0\\ (\vec E-\vec P-\lambda_C \vec v)\cdot \vec v = \vec 0\\ (\vec E-\vec P)\cdot\vec v=\lambda_C \vec v\cdot \vec v\\ \lambda_C = \dfrac{(\vec E-\vec P)\cdot\vec v}{\vec v\cdot \vec v}\\ $$
In your case, $\lambda_C = \dfrac{((4\mathrm i+3\mathrm j)-(2\mathrm i -\mathrm j))\cdot(3\mathrm i+2\mathrm j)}{(3\mathrm i+2\mathrm j)\cdot(3\mathrm i+2\mathrm j)}=\dfrac{(2\mathrm i+4\mathrm j)\cdot(3\mathrm i+2\mathrm j)}{9+4}=\dfrac{6+8}{13}=\dfrac{14}{13}$
Sub $\lambda_C$ into the equation for $L_1$ to get $\vec C$. Then $L_2$ is the line through $\vec C$ and $\vec E$.