# Let \vec a = \langle 3,4 \rangle and \vec b = \langle 1,-1 \rangle. Show that there are scalars s...

## Question:

Let {eq}\vec a = \langle 3,4 \rangle {/eq} and {eq}\vec b = \langle 1,-1 \rangle {/eq}. Show that there are scalars {eq}s {/eq} and {eq}t {/eq} so that {eq}s \vec a + t \vec b = \langle -4,-17 \rangle {/eq}

## Vector:

Vectors can be added or can be scaled by a non-zero number.

That is, for two vectors {eq}\vec a = \langle a_1,a_2 \rangle {/eq} and {eq}\vec b = \langle b_1,b_2 \rangle, \ \vec a +\vec b= \langle a_1+b_1,a_2+b_2 \rangle {/eq} and {eq}k\vec a = k\langle a_1,a_2 \rangle=\langle ka_1,ka_2 \rangle. {/eq}

Given {eq}\vec a = \langle 3,4 \rangle {/eq} and {eq}\vec b = \langle 1,-1 \rangle. {/eq}

Our aim is to find the values of {eq}s {/eq} and {eq}t {/eq} such that {eq}s \vec a + t \vec b = \langle -4,-17 \rangle. {/eq}

Now:

{eq}\begin{align*} s \vec a + t \vec b &=s \langle 3,4 \rangle + t \langle 1,-1 \rangle\\ \\ &= \langle 3s,4s \rangle + \langle t,-t \rangle\\ \\ &= \langle 3s+t,4s-t \rangle=\langle -4,-17 \rangle. \end{align*} {/eq}

By comparing the components, we get: {eq}3s+t=-4 \quad ---(i) \ {/eq} and {eq}4s-t=-17 \quad ---(ii). {/eq}

Now, {eq}(i)\ + (ii) \Rightarrow 7s=-21 \Rightarrow s=-3. {/eq}

Then {eq}3*-3+t=-4 \Rightarrow t=5. {/eq}

Therefore {eq}{\color{Blue} {s=-3}} {/eq} and {eq}{\color{Blue} {t=5}} {/eq}.