[SOLVED] #23 intersecting vector equations of 2 lines

[karush]

View attachment 1504

the calcs are mine, the only way I could see to solve this was to convert these to the $y=mx+b$ line eq and solve simultaneously to the intersection of $(x,y)$

not sure how you take them as they are given and find point P the intersection just as vectors.

also, I tried to input a vector equation of a line in W|A but my syntax didn't produce the correct line. or is there a an input for that.

 

[Ackbach]

Try this, yielding $P=\langle 2,3 \rangle$.

 

[karush]

nice to know that

so $(\lambda,\tau)$ is the intersection?

 

[Ackbach]

No, it is not. $\lambda$ and $t$ are two parameters that, when you change them, help you to traverse the two lines given by the two vector equations you have. To find the point, plug in $\lambda=-1$ into the first equation, or $t=1$ into the second, and the coordinates you get there will be what you need.

 

[MarkFL]

Another approach would be to write the system:

\(\displaystyle 5+3\lambda=-2+4t\)

\(\displaystyle 1-2\lambda=2+t\)

simplify:

\(\displaystyle 4t-3\lambda=7\)

\(\displaystyle t+2\lambda=-1\)

Subtracting 4 times the latter equation from the former, we find:

\(\displaystyle -11\lambda=11\implies\lambda=-1\)

Hence:

\(\displaystyle P=(5+3(-1),1-2(-1))=(2,3)\)