Scalar Product and Angles

  2. Geometry & Trigonometry
  3. Scalar Product and Angles
    4. 30'

In this topic, we will look at finding the angle between vectors in different circumstances. The whole topic revolves around the scalar (or dot) product. Often we are concerned with perpendicular vectors, and the fact that the scalar product equals zero in this case is a hugely important result. Since problems are often set in 3 dimensions, the ability to visualise the situations or draw a quick sketch can help.

Key Concepts

On this page, you should learn about

  • the scalar product of two vectors
  • the angle between two vectors
  • parallel and perpendicular vectors
 

Essentials

The following videos will help you understand all the concepts from this page

Scalar Product & Angles

In the following video, we are going to look at the scalar product and how we can use it to find the angle between two vectors. The scalar product, or the dot product as it is sometimes called, is a way of combining two vectors together that will give us a result that is a scalar (not a vector). The formula for finding the scalar product for 3D can be found in the formula booklet (it is not there for 2D)

2 dimensions

 \({ v }\cdot { w }=\left( \begin{matrix} { v }_{ 1 } \\ { v }_{ 2 } \end{matrix} \right) \cdot \left( \begin{matrix} { w }_{ 1 } \\ { w }_{ 2 } \end{matrix} \right) ={ v }_{ 1 }\cdot { w }_{ 1 }+{ v }_{ 2 }\cdot { w }_{ 2 }\)

3 dimensions

 \(\textbf{ v}\cdot \textbf{w }=\left( \begin{matrix} { v }_{ 1 } \\ { v }_{ 2 } \\ { v }_{ 3 } \end{matrix} \right) \cdot \left( \begin{matrix} { w }_{ 1 } \\ { w }_{ 2 } \\ { w }_{ 3 } \end{matrix} \right) ={ v }_{ 1 }\cdot { w }_{ 1 }+{ v }_{ 2 }\cdot { w }_{ 2 }+{ v }_{ 3 }\cdot { w }_{ 3 }\)

We'll look at the following example

Find the angle between two direction vectors \(\left( \begin{matrix} 1 \\ 2 \\ \sqrt{3} \end{matrix} \right) \) and \(\left( \begin{matrix} -1 \\ 3 \\-2 \end{matrix} \right) \)

​​ Notes from the video

 

Perpendicular Vectors

When two vectors v and w are perpendicular vectors, then \(\textbf{v}\cdot \textbf{w}=0\)

In the following video, we look at the following example

Find a if the following two vectors are perpendicular

\(2\textbf{i}−4\textbf{j}+a\textbf{k}\\ a\textbf{i}+\sqrt{3} \textbf{j}−\textbf{k}\)

Notes from the video

 

Angles between 2 Lines

In the following video we are looking at finding the angle between two lines in 3 dimensional space. The formula for finding the angle between 2 vectors a and b is given in the formula booklet

\(cos\theta =\frac { a\cdot b }{ \left| a \right| \left| b \right| } \)

The example we are going to try and solve is

Find the angle between the lines

\({ L }_{ 1 }:\quad \frac { x-1 }{ -1 } =-y=\frac { z+2 }{ \sqrt { 3 } } \\ { L }_{ 2 }:\quad \frac { x+2 }{ -2 } =\frac { 2y+1 }{ -4 } =z+1\)

Notes from the video

 

Angle between 2 Planes

In the following video we are go to find out how we find the angle between two planes. It is easy to find the angle when the planes are given in Cartesian form as the angle required is the angle between the normals

Whenever we dealing with angles, the formula (found in formula booklet) involving the scalar product is going to be useful

​\(cos\theta =\frac { a\cdot b }{ \left| a \right| \left| b \right| } \)​

Here is the example that will help us understand this topic:

Find the acute angle between the planes

2x + 3y – 4z = 6

and

x - y + 2z = 2

Notes from the video

 

Angle between a Line and a Plane

In the following video we are going to see how we can find the angle between a line and a plane. The key thing is to be able to visualise the problem and to understand what you need to work out. The diagram below should help you see that the angle we need is \(\alpha\) but the angle we can easily find using the normal to the plane is \(\theta\). The video below explains this in detail.

As ever, when we are dealing with angles, the following formula (in the formula booklet) is going to be useful

​\(cos\theta =\frac { a\cdot b }{ \left| a \right| \left| b \right| } \)​

The example we are going to look at is as follows:

Find the acute angle between the line and plane

\(-x\ =\ \frac { y-5 }{ 2 } \ =\ 2z-8\)

and

3x - y + z = 8

Notes from the video

 
 

Summary

Print from here

text

Test Yourself

The following quiz tests your knowledge and understanding of the scalar product, angles between vectors and angles between lines


START QUIZ!

The following quiz tests your knowledge and understanding of angles between

  1. Two Lines
  2. Two Planes
  3. A Line and a Plane

START QUIZ!

text

Exam-style Questions

Question 1

A line \({ L }_{ 1 }\) passes through A(2,0,-3) and B(4,3,2).

a) Find the equation of the line \({ L }_{ 1 }\)

A second line \({ L }_{ 2}\) has equation \(\textbf{r}=\left( \begin{matrix} 2 \\ 3 \\ 5 \end{matrix} \right) +\lambda \left( \begin{matrix} 1 \\ -4 \\ k \end{matrix} \right) \)

b) Given that \({ L }_{ 1 }\) and \({ L }_{ 2 }\)are perpendicular, find k.

Hint

Full Solution

 

 

Question 2

\(\overrightarrow { AB }\) and \(\overrightarrow { AC }\) are two vectors such that \(\overrightarrow { AB } =\left( \begin{matrix} 3 \\ -1 \\ 2 \end{matrix} \right) \) and \(\overrightarrow { AC } =\left( \begin{matrix} 2 \\ 0 \\ 1 \end{matrix} \right) \)

Find \(\hat { BAC } \) to the nearest degree.

Hint

Full Solution

 

 

Question 3

The angle between the line \({ L }_{ 1 }\) and \({ L }_{ 2 }\) is \(\frac{\pi}{2}\).

\( { L }_{ 1 }:\quad \frac { x+2 }{ 3 } =2y+1=\frac { 5-z }{ 2 } \)

\( { L }_{ 2 }: \quad x =\frac { y-2}{ 3} =kz \)

Find k.

Hint

Full Solution

 

 

Question 4

Find the angle between the planes \({ \Pi }_{ 1 }\) and \({ \Pi }_{ 2 }\) to the nearest degree.

\({ \Pi }_{ 1 }: 2x-3y+z=0\)

\({ \Pi }_{ 2 }: x+2y+5z=-4\)

Hint

Full Solution

 

 

Question 5

Find the value of x for which the vectors \(\left( \begin{matrix} sinx \\ \sqrt{3} \\ 0 \end{matrix} \right) \) and \(\left( \begin{matrix} 4cosx \\-1\\ 2 \end{matrix} \right) \)are perpendicular, \(0\le x\le \frac { \pi }{ 2 } \).

Hint

Full Solution

 

 

Question 6

OABC is a parallelogram.

\(\overrightarrow { OA } =\textbf{a}\) \(\overrightarrow { OB } =\textbf{b}\) \(\overrightarrow { OC } =\textbf{a}+\textbf{b}\)

Given that \((\textbf{ a }+\textbf{ b })\cdot (\textbf{ a }-\textbf{ b })=0\) what can you conclude

 

Hint

Full Solution

 

 

Question 7

ACB is a right-angled triangle

\(\overrightarrow { CB } = \textbf {a }\) \(\overrightarrow { AC } = \textbf {b }\)

a) Write \(\overrightarrow { AB } \) in terms of a and b

b) Find \(\textbf{ a }\cdot \textbf{ b }\)

c) Show that \({ \left| \textbf { a+b } \right| }^{ 2 }={ \left| \textbf { a } \right| }^{ 2 }+{ \left| \textbf{ b } \right| }^{ 2 }\) and hence prove Pythagoras' Theorem.

Hint

Full Solution

 

 

MY PROGRESS

How much of Scalar Product and Angles have you understood?