*IBV6 Find vector eq of line passing thru (–1, 4),(3, –1). in form r = p + td

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SUMMARY

The vector equation of the line passing through the points (–1, 4) and (3, –1) is expressed as $$r = (3, -1) + t(4, -5)$$, where $$t \in \mathbb{R}$$. The position vector is identified as (3, -1), and the direction vector is calculated as (4, -5) by subtracting the coordinates of the two points. This formulation provides a clear representation of the line in vector form.

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karush
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Find a vector equation of the line passing through
$(–1, 4)$ and $(3, –1)$.
Give answer in the form $$r = p + td$$,
where $$t \in {R}$$

position vector would be $$(3, -1) = p$$

direction vector would be (3+1,-1-4) = (4,-5) = d

so $$r=(3,-1)+t(4,-5)$$
 
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