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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.

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  • $\begingroup$ 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. $\endgroup$ – Nicolas Feb 4 '16 at 12:12
  • $\begingroup$ 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 :/ $\endgroup$ – Tom Symonds Feb 4 '16 at 12:17
  • $\begingroup$ 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. $\endgroup$ – Nicolas Feb 4 '16 at 12:30
  • $\begingroup$ 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. $\endgroup$ – pjs36 Feb 4 '16 at 12:31
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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 $\vec E$.

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  • $\begingroup$ Omg thank you so much!! That was the best explanation I've seen yet. Thank you xxx $\endgroup$ – Tom Symonds Feb 6 '16 at 13:33
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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 $b_p$.

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  • $\begingroup$ 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...? $\endgroup$ – Tom Symonds Feb 4 '16 at 12:18
  • $\begingroup$ Use the definition of a scalar product and 90 degree rotation matrix and the same result will follow. $\endgroup$ – user296856 Feb 4 '16 at 12:21

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