# CALC III: Vectors problem?

Let a = < 5, 3, -4 > and b = <1, 0, 2>. Show that there are scalars s and t so that sa + tb = < -13, -9, 16 >

Find

s=

t=

Relevance
• Plug in the vectors a and b

s<5, 3, -4> + t<1, 0, 2> = <-13, -9, 16>

Add the components and equate them to the components of the last vector

5s + t = -13

3s = -9

-4s + 2t = 16

Solve for s and t, use the 2nd equation, s = -3. Plug this into the first equation:

(-15) + t = -13, solve for t = 2

So, plug in the scalars s = -3 and t = 2 for the last equation. If the equation holds true, then there are scalars s and t for which sa + tb = <-13, -9, 16>

Plug in: -4(-3) + 2(2) = 12 + 4 = 16

This holds true as 16 is the last component of the vector <-13, -9, 16>.

Hope this helps.

• a = 5i + 3j -4k

b = i + 2k

if you simply added these vectors together you would get 6i + 3j - 2k

So obviously there are scalars s & t that will give the answer of -13i - 9j + 16k

s(5i + 3j - 4k) + t(i + 2k) = -13i - 9j + 16k

solve for the scalars