Vector Algebra: Finding the Angle Between Two Vectors

In summary, we are given four non-zero vectors $\vec{a},\vec{b}, \vec{c}$ and $\vec{d}$ in which $\vec{a},\vec{b}$ and $\vec{c}$ are coplanar but not collinear pair by pair, and $\vec{d}$ is not coplanar with vectors $\vec{a},\vec{b}$ and $\vec{c}$. The angles between the vectors are given by $(\widehat{\vec{a}\vec{b}})=(\widehat{\vec{b}\vec{c}})=\frac{\pi}{3}, (\widehat{\vec{d}\vec{a}})=\alpha, (\
  • #1
Saitama
4,243
93
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!
 
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  • #2
Pranav said:
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 $\vec{a}=\vec{c}$.

Pranav said:
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 $\hat{e}=\vec{e}/|\vec{e}|$.)
 
  • #3
Evgeny.Makarov said:
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. :)
 
  • #4
Pranav said:
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.

Pranav said:
Not obvious to me.
Consider a hexagon.

200px-Regular_polygon_6_annotated.svg.png


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 $AB=OC$.
 
  • #5
Evgeny.Makarov said:
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! :)
 
  • #6
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:
$$\hat c = \hat b - \hat a$$
$\blacksquare$
 
  • #7
I like Serena said:
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. :)
 

FAQ: Vector Algebra: Finding the Angle Between Two Vectors

What is vector algebra?

Vector algebra is a branch of mathematics that deals with the manipulation and analysis of vectors, which are quantities that have both magnitude and direction. It involves performing operations such as addition, subtraction, and multiplication on vectors to solve problems in various fields, including physics, engineering, and computer graphics.

What are the basic operations in vector algebra?

The basic operations in vector algebra are addition, subtraction, and scalar multiplication. Addition involves combining two or more vectors to form a resultant vector, while subtraction is the inverse operation of addition. Scalar multiplication involves multiplying a vector by a scalar, which is a real number, to change its magnitude without altering its direction.

How do I represent vectors in vector algebra?

Vectors in vector algebra are typically represented using symbols that have both magnitude and direction. The magnitude of a vector is denoted by its length, while its direction is represented by an arrow pointing in the direction of the vector. Vectors can also be represented using Cartesian coordinates, where the x, y, and z components indicate the magnitude of the vector in each dimension.

What are some common applications of vector algebra?

Vector algebra has many applications in the real world, including navigation, motion analysis, and force calculations. In navigation, vectors are used to represent the direction and magnitude of an object's movement or position. In motion analysis, vectors are used to describe the velocity and acceleration of an object. In force calculations, vectors are used to determine the resultant force acting on an object with multiple forces acting on it.

How can I solve vector algebra problems?

To solve vector algebra problems, you need to have a good understanding of the basic operations and properties of vectors. You should also be familiar with vector notation and how to represent vectors in different coordinate systems. It is important to carefully read the problem and identify the given information, as well as the unknowns that need to be solved for. Then, you can use the appropriate operations and equations to manipulate the vectors and find the solution.

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