How to find the vector, cosine and equation of two intersecting lines.

Question

I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. (I haven’t learned any of this, but I am trying to become familiar with it.) I found a website full of problems of the content we will be learning, but it doesn't have the answers. I need a little guidance on how to do this problem.

I’m going to do my best to talk through each part and tell you how I think it should be solved!

A line $L_1$ passes through points $P(-1, 6, -1)$ and $Q(0, 4, 1)$.

a. (i) Show that vector $PQ = (1, -2, 2)$.
So, I can see that if you take the $x$, $y$, and $z$ of $Q$, and subtract the $x$, $y$, and $z$ of $P$, then you get $PQ$. But, is there some sort of formula for this that you would use to answer this question?

(ii) Hence, write down an equation for $L_1$ in the form $r = a+tb$.
Would this be: $r= (-1, 6, -1) + t (0, 4, 1)$? (I just plugged in $P$ for $a$ and $Q$ for $b$).

b. A second line $L_2$ has equation $r = (4, 2, -1) + s (3, 0, -4)$.
Find the cosine between vector $PQ$ and $L_2$.
I’m super confused on how you could find the cosine of this when all you have is some points? Or, if I draw this out somehow would I find the sides to do the Law of Cosines?

c. The lines $L_1$ and $L_2$ intersect at the point $R$. Find the coordinates of $R$.
Would I use a midpoint formula to discover this? Since $R$ would be in the middle of both lines?

Answer 1

For a, (i) and (ii) you are correct. Just subtract components to get the span vector.

For b. you need to use dot product rule that connects the magnitude of two direction vectors and the angle between them. The dot product of $(1, -2, 2)$ and $(3, 0, -4)$ is related to the cosine of the angle between the vectors

$$ \boldsymbol{a} \cdot \boldsymbol{b} = \| \boldsymbol{a} \| \| \boldsymbol{b} \| \cos \theta $$

_{A direction vector is the part of the line equation that multiplies the parameter.}

For c. you need to find $t$ and $s$ when $r$ are equal. You try to solve

$$ \boldsymbol{R} = \pmatrix{-1 \\ 6 \\ -1} + t\,\pmatrix{0 \\ 4 \\ 1} = \pmatrix{4 \\ 2 \\ -1} + s\,\pmatrix{3 \\ 0 \\-4} $$

Answer 2

For (ai), you're correct. For instance you want to find $\vec{PQ}$, with given coordinates $P(p_1,p_2,\cdots,p_n)$ and $Q(q_1,q_2,\cdots,q_n)$, $\vec{PQ}=(q_1-p_1,q_2-p_2,\cdots,q_n-p_n)$.

However, for (aii), an equation for $L$ in the form $r = a+tb$ is given by $r=(-1,6,-1)+t(1,-2,2)$. To see this, you may draw a picture. Start at point $P(-1,6,-1)$, and traverse along ($t$) the line $L$ in the direction of $\vec{PQ}=(1,-2,2)$.

For (b), we have another line $L_2:r = (4, 2, -1) + s (3, 0, -4)$. This line is in the direction $(3, 0, -4)$. By the formula $$\cos\theta=\dfrac{A\cdot B}{|A||B|}$$which gives you the cosine of two lines $A,B$ (here "$\cdot$" is the dot product), you have $$\cos\theta=\dfrac{3-8}{3\cdot5}=\dfrac{-5}{15}=-\dfrac{1}{3}.$$ You may also do it by Law of Cosines (I guess), but I don't think it would be easy with just vectors.

For (c), yes as ja72 has noted, you have to find $s,t$ such that $L_1=L_2$. You can do it by equating coordinates: $$\begin{cases}-1+t&=4+3s\\6-2t&=2\\-1+2t&=-1-4s\end{cases}$$

Pick any two equations to solve, and verify with the third equation.