# Finding perpendicular vectors using dot product

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.

• If $(x,y)$ is the cartesian coordinates of your directon vector, then a direction vector $(u,v)$ of the perpendicular line is such that $(a,b)\cdot (u,v)=0$, where $\cdot$ denotes the dot product. Feb 4, 2016 at 12:12
• Ok thanks. That makes sense, so I've come up with the equation 3x+ 2y = 0, however I cant solve an equation that has two variables :/ Feb 4, 2016 at 12:17
• I am not sure about your equation : the direction vector of line 1 is $(2+3\lambda,-1+2\lambda)$. So if $(u,v)$ is a direction vector of the perpendicular line, then $0=(2+3\lambda,-1+2\lambda)\cdot(u,v)=(2+3\lambda)u+(-1+2\lambda)v$. You want to find $u$ and $v$ : take ANY values that can do the job, for example $u=1$ and $v=-\frac{2+3\lambda}{-1+2\lambda}$. Then the equation of the perpendicular line (with direction $(u,v)$) is given by $y=ax+b$ where $a=v/u$ is the direction coefficient, and $b$ is found by taking $(x,y)=(4\mathrm{i}+3\mathrm{j})$. I let you try it. Feb 4, 2016 at 12:30
• Well, you need two variables (in general) to describe a line; if you somehow did get a single $(x,y)$ pair, you'd have a point instead. Feb 4, 2016 at 12:31

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 $\vec E$.

• Omg thank you so much!! That was the best explanation I've seen yet. Thank you xxx Feb 6, 2016 at 13:33

If two cartesian lines are perpendicular, the product of their slopes are -1.

So if you have $y = kx + b$

Then $y_p = -\frac{1}{k}x+b_p$, assuming $k$ is not zero

Plug in the position vector to find $b_p$.

• Thanks for your help, but this question specifically states that the dot product (also called the scalar product) must be used. Not sure how to do this...? Feb 4, 2016 at 12:18
• Use the definition of a scalar product and 90 degree rotation matrix and the same result will follow. Feb 4, 2016 at 12:21