# Vector algebra problem

#### Pranav

##### Well-known member
Problem:
Given four non-zero vectors $\vec{a},\vec{b},\vec{c}$ and $\vec{d}$, the vectors $\vec{a},\vec{b}$ and $\vec{c}$ are coplanar but not collinear pair by pair and $\vec{d}$ is not coplanar with vectors $\vec{a},\vec{b}$ and $\vec{c}$ and $(\widehat{\vec{a}\vec{b}})=(\widehat{\vec{b}\vec{c}})=\frac{\pi}{3}, (\widehat{\vec{d}\vec{a}})=\alpha, (\widehat{\vec{d}\vec{b}})=\beta$, then prove that $(\widehat{\vec{d}\vec{c}})=\arccos(\cos\beta-\cos\alpha)$.

$\widehat{\vec{a}\vec{b}}$ denotes the angle between two vectors.

Attempt:
I have the following:
$$\hat{a}\cdot\hat{b}=\hat{b}\cdot\hat{c}=\frac{1}{2}$$
$$\hat{d}\cdot\hat{a}=\cos\alpha$$
$$\hat{d}\cdot\hat{b}=\cos\beta$$
Subtracting the above two equations, I get
$$\hat{d}\cdot (\hat{b}-\hat{a})=\cos\beta-\cos\alpha$$
I somehow need to show that $\hat{c}=\hat{b}-\hat{a}$ but I don't see how. I notice that $\vec{b}$ is the angle bisector of $\vec{a}$ and $\vec{c}$ but I am not sure if that helps.

Any help is appreciated. Thanks!

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Given four non-zero vectors $\vec{a},\vec{b},\vec{c}$ and $\vec{d}$, the vectors $\vec{a},\vec{b}$ and $\vec{c}$ are coplanar but not collinear pair by pair and $\vec{d}$ is not coplanar with vectors $\vec{a},\vec{b}$ and $\vec{c}$ and $(\widehat{\vec{a}\vec{b}})=(\widehat{\vec{b}\vec{c}})=\frac{\pi}{3}, (\widehat{\vec{d}\vec{a}})=\alpha, (\widehat{\vec{d}\vec{b}})=\beta$, then prove that $(\widehat{\vec{d}\vec{c}})=\arccos(\cos\beta-\cos\alpha)$.

$\widehat{\vec{a}\vec{b}}$ denotes the angle between two vectors.
The directed angle? Because the claim does not necessarily hold if $\vec{a}=\vec{c}$.

I somehow need to show that $\hat{c}=\hat{b}-\hat{a}$
If $\vec{a}$ and $\vec{c}$ lie on different sides of $\vec{b}$, then the fact that $\hat{c}=\hat{b}-\hat{a}$ is obvious from the sketch of the plane. (I assume $\hat{e}=\vec{e}/|\vec{e}|$.)

#### Pranav

##### Well-known member
The directed angle? Because the claim does not necessarily hold if $\vec{a}=\vec{c}$.
I am not sure, I don't know the latex code for the symbol used to denote the angle between two vectors. The symbol used in my practice sheet is a triangle without the base. \widehat looked similar so I used that instead.
If $\vec{a}$ and $\vec{c}$ lie on different sides of $\vec{b}$, then the fact that $\hat{c}=\hat{b}-\hat{a}$ is obvious from the sketch of the plane.
Not obvious to me.

(I assume $\hat{e}=\vec{e}/|\vec{e}|$.)
Yes.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
I am not sure, I don't know the latex code for the symbol used to denote the angle between two vectors. The symbol used in my practice sheet is a triangle without the base. \widehat looked similar so I used that instead.
It does not matter how the angle is denoted and whether you can reproduce it in LaTeX. It is important that you know the meaning behind the notation. At a minimum, I suggest you consider the case where $\vec{a}=\vec{c}$ and both form a $60^\circ$ angle with $\vec{b}$. Then the claim is not necessarily true. The best thing would be to go back in your sources and find the definition of this notation. Then you will know if the problem is correct or if it is sloppily stated.

Not obvious to me.
Consider a hexagon.

Let the center be denoted by $O$, the top vertex by $A$, and, going clockwise, the following vertices be denoted by $B$ and $C$. All angles in triangles are $60^\circ$, so angle $BOC$ equals angle $OBA$. They are alternating angles for $AB$ and $OC$, so these segments are parallel. It is also easy to see that $AB=OC$.

#### Pranav

##### Well-known member
It does not matter how the angle is denoted and whether you can reproduce it in LaTeX. It is important that you know the meaning behind the notation. At a minimum, I suggest you consider the case where $\vec{a}=\vec{c}$ and both form a $60^\circ$ angle with $\vec{b}$. Then the claim is not necessarily true. The best thing would be to go back in your sources and find the definition of this notation. Then you will know if the problem is correct or if it is sloppily stated.
I never saw that kind of notation being used during the class. It is the first time I have seen the notation. The practice sheet (not provided by my teacher) is also silent and has no clarification so I took a guess that its the angle between the two vectors. Also, I have never heard anything like "directed angle" before.

I have figured out $\hat{c}=\hat{b}-\hat{a}$, thanks a lot Evgeny.Makarov!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hey Pranav!

Here's an alternative method.

Since $\vec a, \vec b, \vec c$ are coplanar and pairwise independent, there have to be unique numbers $\lambda$ and $\mu$, such that:
$$\hat c = \lambda \hat a + \mu \hat b$$

Combine with your first two dot products to find:
\begin{array}{}
\hat c \cdot \hat b &=& \lambda \hat a \cdot \hat b + \mu \hat b \cdot \hat b &=& \frac 1 2 \lambda + \mu &=& \frac 1 2 \\
\hat c \cdot \hat c &=& \lambda^2 \hat a^2 + 2\lambda\mu\hat a \cdot \hat b + \mu^2 \hat b^2 &=& \lambda^2 + \lambda\mu + \mu^2 &=& 1
\end{array}

Solve to find:
$$\lambda=-1, \mu=1$$
Therefore:
$$\hat c = \hat b - \hat a$$
$\blacksquare$

#### Pranav

##### Well-known member
Hey Pranav!

Here's an alternative method.

Since $\vec a, \vec b, \vec c$ are coplanar and pairwise independent, there have to be unique numbers $\lambda$ and $\mu$, such that:
$$\hat c = \lambda \hat a + \mu \hat b$$

Combine with your first two dot products to find:
\begin{array}{}
\hat c \cdot \hat b &=& \lambda \hat a \cdot \hat b + \mu \hat b \cdot \hat b &=& \frac 1 2 \lambda + \mu &=& \frac 1 2 \\
\hat c \cdot \hat c &=& \lambda^2 \hat a^2 + 2\lambda\mu\hat a \cdot \hat b + \mu^2 \hat b^2 &=& \lambda^2 + \lambda\mu + \mu^2 &=& 1
\end{array}

Solve to find:
$$\lambda=-1, \mu=1$$
Therefore:
$$\hat c = \hat b - \hat a$$
$\blacksquare$
Thanks ILS! That is also a nice way to solve the problem.