Angle between vectors $\vec a + \vec b$ and $\vec c$?

Question

If there are three vector $\vec a$, $\vec b$ and $\vec c$ provided, what is the angle between $\vec a + \vec b$ and $\vec c$? I understand how to calculate for angle between $\vec a$ and $\vec b$, $\vec a$ and $\vec c$ and $\vec b$ and $\vec c$ but what does the angle between $\vec a + \vec b$ and $\vec c$ mean?

Answer 1

Just $$\cos(\widehat{\vec{a}+\vec{b},\vec{c}})=\frac{(\vec{a}+\vec{b})\vec{c}}{|\vec{a}+\vec{b}||\vec{c}|}.$$

Here, $|\vec{a}+\vec{b}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+2\vec{a}\vec{b}}$ and $(\vec{a}+\vec{b})\vec{c}=\vec{a}\vec{c}+\vec{b}\vec{c}.$

Answer 2

You compute the sum of $a+b$, that is a vector, if it helps, you can give a new name, $d$.

Now you can compute the angle between $d$ and $c$ using the formula that you are familiar with.

Answer 3

We have that $\vec d=\vec a+\vec b$ then

$$\cos (\theta)=\frac{\vec d \cdot \vec c}{|\vec d||\vec c|}$$

Answer 4

Using the definition of the scalar product we get $$ (a+b)\cdot c = \lVert a+b \rVert \lVert c \rVert \cos \angle(a+b,c) $$ where $\lVert v \rVert = \sqrt{v\cdot v}$.

For non-zero $a+b$ and $c$ we can solve for the angle $$ \angle(a+b,c) = \arccos \frac{(a+b)\cdot c}{\lVert a+b \rVert \lVert c \rVert} $$

Answer 5

The important concept to keep in mind here is that when you add two vectors, $\vec a$ and $\vec b$, the result is just another vector. It has a magnitude, a direction (if its magnitude is positive), and components just like any other vector.

We may write the expression $\vec a + \vec b$ in order to name this vector, but that does not make it into any new kind of object. The formulas that apply to other vectors still apply.

To find the angle between $\vec a + \vec b$ and $\vec c,$ take the formula you would use for the angle between any two vectors, put $\vec a + \vec b$ in the places where the first vector occurs in the formula, and put $\vec c$ in the places where the second vector occurs. Look at other answers for details of what happens after you do this.