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Problem 46 Hard Difficulty

Suppose that $ a $ and $ b $ are nonzero vectors that are not parallel and $ c $ is any vector in the plane determined by $ a $ and $ b $. Give a geometric argument to show that $ c $ can be written as $ c = sa + tb $ for suitable scalars $ s $ and $ t $. Then give an argument using components

Answer

Let $\mathbf{a}=\left\langle a_{1}, a_{2}\right\rangle, \mathbf{b}=\left\langle b_{1}, b_{2}\right\rangle$ and $\mathbf{c}=\left\langle c_{1}, c_{2}\right\rangle .$ So $\mathbf{c}= s \mathbf{a}+t \mathbf{b}$ means $\left\langle c_{1}, c_{2}\right\rangle= s\left\langle a_{1}, a_{2}\right\rangle+ t\left\langle b_{1}, b_{2}\right\rangle=$ $\left\langle s a_{1}, s a_{2}\right\rangle+\left\langle t b_{1}, t b_{2}\right\rangle=\left\langle s a_{1}+t b_{1}, s a_{2}+t b_{2}\right\rangle .$ That is $ c_{1}=s a_{1}+t b_{1}$ and $c_{2}=s a_{2}+t b_{2} .$ The argument here is that these two equations form a system of linear equations which has a unique solution for the values of $s$ and $t$ because $\left\langle a_{1}, a_{2}\right\rangle$ and $\left\langle b_{1}, b_{2}\right\rangle$ are not parallel to one another (i.e., they are not multiple of one another).

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Video Transcript

So let's suppose that A. And B are non zero vectors that are not parallel to see and see as any vector in the plane. So we want to give a geometric argument to show that C can be written as um the sum of a plus B. With scalar multiples, S and T. So we can use the argument, give the argument using components. So we know that C is going to be C sub X. He said why? And we want to show that's equal to if we have a plus or a X. A. Y. Name. Plus. Yes, do I? Well we noticed that we could add these apps, so we get A X plus bx and then A Y plus B. Y. But they had scalar multiples. So what we see in this case is that we would have um S A N T. B. So S here T. Here S here T. Here. So when we do this, we see that we can have any scalar multiples to get to be ex well, we know if we just took out the S. And the T. And we take A. X. Plus bx and multiplied by some scalar multiple, then that's going to end up giving us C. X. And the same thing holds for the white components, so we see that this is going to hold.