What is solution to this vector problem?

Question

Problem 38

Suppose $|\vec{u}|=3$, $|\vec{v}|=5$ and $\vec{u}\cdot\vec{v}=-2$. Calculate $(\vec{u}-\vec{v})\cdot(2\vec{u}+3\vec{v})$.

I am only familiar with this formula $\vec{a}\cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\theta)$. But it doesn't help much.

Answer 1

If the $\circ$ denotes the dot product: $$(\vec u-\vec v)\circ(2\vec u+3\vec v)$$ Let's call $\vec w:=\vec u-\vec v$, then: $$\vec w\circ(2\vec u+3\vec v)$$ Because the dot product is linear: $$\vec w\circ(2\vec u+3\vec v)=\vec w\circ(2\vec u)+\vec w\circ(3\vec v)=2(\vec w\circ\vec u)+3(\vec w\circ\vec v)$$ Now substitute back $w$: $$2(\vec w\circ\vec u)+3(\vec w\circ\vec v)=2([\vec u-\vec v]\circ\vec u)+3([\vec u-\vec v]\circ\vec v)$$ We can use the linearity again: $$2([\vec u-\vec v]\circ\vec u)+3([\vec u-\vec v]\circ\vec v)=2(\vec u\circ\vec u-\vec v\circ\vec u)+3(\vec u\circ\vec v-\vec v\circ\vec v)$$ Now we can use that $\vec a\circ\vec b=\vec b\circ\vec a$, and $\vec a\circ\vec a=|\vec a||\vec a|\cos(0)=|\vec a|^2$: $$2(\vec u\circ\vec u-\vec v\circ\vec u)+3(\vec u\circ\vec v-\vec v\circ\vec v)=2(|\vec u|^2-\vec v\circ\vec u)+3(\vec u\circ\vec v-|\vec v|^2)=2|\vec u|^2-2\vec v\circ\vec u+3\vec u\circ\vec v-3|\vec v|^2=2|\vec u|^2+\vec u\circ\vec v-3|\vec v^2|=2*3^2-2-3*5^2=-59$$

Answer 2

HINT

Use the following basic properties of dot product

• $\vec a \cdot (\vec b+\vec c)=\vec a \cdot \vec b + \vec a \cdot \vec c$

• $\vec a \cdot \vec a =|\vec a|^2$

• $\vec a \cdot \vec b = \vec b \cdot \vec a $

Answer 3

The dot product is distributive and commutative so we can write $(\overrightarrow{u}-\overrightarrow{v}).(2\overrightarrow{u}+3\overrightarrow{v})=2(\overrightarrow{u}.\overrightarrow{u})+(\overrightarrow{u}.\overrightarrow{v})-3(\overrightarrow{v}.\overrightarrow{v})$

Now we use the property that for any vector $A$ we have $A.A=|A|^{2}$ and we can now substitute in the given values hence $2(\overrightarrow{u}.\overrightarrow{u})+(\overrightarrow{u}.\overrightarrow{v})-3(\overrightarrow{v}.\overrightarrow{v})=2|\overrightarrow{u}|^{2}+\overrightarrow{u}.\overrightarrow{v}-3|\overrightarrow{v}|^{2}=2\times3^{2}-2-3\times5^{2}=-59$.