#23 intersecting vector equations of 2 lines

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SUMMARY

The discussion focuses on solving the intersection of two lines represented by vector equations. The primary method involves converting the equations to the slope-intercept form $y=mx+b$ and solving them simultaneously. The intersection point, denoted as $P$, is calculated by substituting parameters $\lambda$ and $t$ into the vector equations, ultimately yielding the coordinates $P=\langle 2,3 \rangle$. The discussion also highlights the use of a system of equations to derive the intersection point accurately.

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  • Understanding of vector equations and their representation.
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karush
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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.
 
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Try this, yielding $P=\langle 2,3 \rangle$.
 
nice to know that

so $(\lambda,\tau)$ is the intersection?
 
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.
 
Another approach would be to write the system:

$$5+3\lambda=-2+4t$$

$$1-2\lambda=2+t$$

simplify:

$$4t-3\lambda=7$$

$$t+2\lambda=-1$$

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

$$-11\lambda=11\implies\lambda=-1$$

Hence:

$$P=(5+3(-1),1-2(-1))=(2,3)$$
 

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