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# (a) Draw the vectors $a = \langle 3, 2 \rangle$, $b = \langle 2, -1 \rangle$, and $c = \langle 7, 1 \rangle$.(b) Show, by means of a sketch, that there are scalars $s$ and $t$ such that $c = sa + tb$.(c) Use the sketch to estimate the values of $s$ and $t$.(d) Find the exact values of $s$ and $t$.

## a) Though they don't have to, we can draw all vectorsinitiating from the origin. In that case, a will terminateat the point $(3,2),$ b at $(2,-1)$ and $\mathbf{c}$ at $(7,1)$ .b) As you can see sa is some scalar multiple of the avector and $t \mathrm{b}$ is some scalar multiple of the b vectorand they form a parallelogram where the c vector isthe diagonal of the parallelogram. Thus $\mathbf{c}=s \mathbf{a}+t \mathbf{b}$for some scalars $s$ and $t .$c) Now if $s$ is indeed 4$/ 3$ and $t=3 / 2,$ then $s \mathbf{a}+t \mathbf{b}=$$\frac{4}{3}\langle 3,2\rangle+\frac{3}{2}\langle 2,-1\rangle=\left\langle 4, \frac{8}{3}\right\rangle+\left\langle 3,-\frac{3}{2}\right\rangle=\left\langle 7, \frac{7}{6}\right\rangled)$$s=\frac{9}{7} \text { and } t=\frac{11}{7}$\$

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So let's start by drawing the vectors. Um Let's get the actual X axis and Z axis and grid back. So we see that We want to draw equals 32. So this is the A vector right here. And then we want to graph the B vector which is going to be 2 -1. And lastly we want to graph the c vector which is 71. And then by means of a sketch, we want to show that there are scholars, S and T. Such that as plus A. Equals or a C equals S. A plus TB. Um So we noticed that if for example, we drew this sketch right here, um parallel and do this one right here. It would create this vector right here, which is just a scaler multiple of this larger vector. So we see that in that case we would be able to scale it such that some of the vectors would equal the larger vector.

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