0
$\begingroup$

Given: $\vec{a} (1; -2), ~\vec{b} (-3; 1), ~\vec{c} (2; -3)$

How to find: $(3a-b)(b+2c)$ - ?

I know: $\vec{a} \cdot \vec{b}$ and $(\vec{a}+\vec{b}) \cdot \vec{c}$, but $(3a-b)(b+2c)$ is scalar product of two new vectors?

3 Answers 3

2

think about each term separately. You have $3\vec{a}$ which is the scalar multiple of $\vec{a}$ hence you have $3\vec{a} = (3,-6)$

Then you take the difference $3\vec{a}-\vec{b}$ and you have $$(3,-6) - (-3,1) = (6,-7)$$

Similarly $2\vec{c} = (4,-6)$ and hence $\vec{b}+2\vec{c}$ is $$(-3,1) + (4,-6) = (1,-5)$$

Then you take the dot product and have $$(6,-7)\cdot (1,-5) = 6 + 35 = 41$$

  • 0
    $(3,-6) - (-3,1) = (9,-7)$ - ? – 2017-01-25
  • 0
    ehm... that's embarassing – 2017-01-25
  • 0
    @divisor I've corrected it now – 2017-01-25
  • 0
    No, I do many mistakes, too :) – 2017-01-25
  • 0
    Thank you, I checked my solution :) – 2017-01-25
0

Do you know that the scalar product is bilinear? E.g. $(u+v)w = uw + vw$ for all $u,v,w$... If you use that you will manage. :) Otherwise you could just compute $3a-b$ and $b+2c$ as all vectors are given.

0

You will find all the tools you need here Vector Calculus.