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Let $\bf{u} \in \mathbb{R}^2$ be the vector $\begin{pmatrix}3\\-4 \end{pmatrix}$. Let $l$ be the line with equation

${\bf{u}} \cdot {\bf{v}} = 50$

a) Express $l$ in parametric form

b) Write the equation of the line perpendicular to $l$ that passes through the origin using vector notation

c) Find the point of intersection of these two lines. Hence or otherwise find the distance from $l$ to the origin

Please can someone explain this to me step by step? This is a past exam question. I have the answers in front of me but there is no explanation of how you get to them, which is not much help.

If it helps:

a) $\begin{pmatrix}6\\-8 \end{pmatrix} + \lambda\begin{pmatrix}4\\3 \end{pmatrix}$

b) $\begin{pmatrix}4\\3 \end{pmatrix} \cdot {\bf{v}} = 0$

c) $(6,-8)$. $10$.

math_apprentice
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  • Have you tried searching the site (via the search box in the upper right corner of the page)? The on-site search engine isn't that great, but these are common enough questions that I think you could find several examples of each type fairly quickly. –  Jun 01 '17 at 20:41
  • I have tried searching the site but most of the answers are using two points, etc. and not the dot product. I think that's what has thrown me as I've never seen a problem phrased in this way before. – math_apprentice Jun 01 '17 at 20:47
  • Let $\mathbf v = (x,y)$. Then $\mathbf u\cdot\mathbf v = 50$ is just a more compact way of writing $$3x-4y=50$$ –  Jun 01 '17 at 20:47

1 Answers1

2

Solutions

$\textbf{(a)}$ From $3x-4y=50$, let $x=t$ (where $t$ is called the parameter). Then $y=\frac 34t-\frac {25}{2}$. So $$(x,y) = \left(t,\frac 34t-\frac {25}2\right) = t\left(1,\frac 34\right)+\left(0,-\frac {25}2\right)$$ If you wanted to get rid of the fraction in the first vector, then define a new parameter $\lambda = \frac 14t$. Then we have $$(x,y) = \lambda\left(4,3\right)+\left(0,-\frac {25}2\right)$$ There's no real real to try to get rid of the fraction in the second vector in this equation, but if you just wanted to verify that the equation given in your solutions is equivalent to this one you could define a new variable again: $\alpha = \lambda -\frac 32$.

$\textbf{(b)}$ You want a vector perpendicular to $(3,-4)$. There are infinitely many, but a little trick that works for 2D vectors is to swap the coordinates and negate one. So $(4,3)$ or $(-4,-3)$. That'll describe the direction of your new line (technically it's orthogonal to your new line, but that's how it goes in $\mathbf u\cdot\mathbf v = d$ form). The fact that it passes through the origin means that $\mathbf v = (0,0)$ should be a solution to $(4,3)\cdot\mathbf v = d$. So $d=0$ and you get your equation.

$\textbf{(c)}$ Plug the parametric coordinates from part (a) into the equation found in part (b) and solve for the parameter. Then plug the parameter back into your parametric equation from part (a) to get the point of intersection. The distance from the origin is then calculable using the Pythagorean theorem.

Further Reading

The following are links to answers I've given on similar past questions. They provide a bit more information than I can and have in this one: