Vector Basics

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Vector Basics

This section covers what vectors are and how to add them together.

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What are Vectors?

A vector contains both size and direction, for example a velocity of 5 m/s5text{ m/s} on a bearing of 120°120degree

As shown on the right, vectors are drawn as lines with arrowheads on them.

The magnitude (size) of the vector is expressed by the length of the line, these are sometimes drawn to scale.

The direction of the vector is shown by the direction of the arrow.

Note: Vectors represented by a single letter are either written using bold font, aboldsymbol{a}, or with an underline, aunderline{a}.

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Finding Resultant Vectors

Vectors can be added together by drawing them, then the single vector that goes from the start of the first vector to the end of the final vector is known as the resultant vector.

We can see that to get to the end point of rboldsymbol{r}, we need to go along the vector aboldsymbol{a} then the vector bboldsymbol{b}.

Thus, r=a+bboldsymbol{r}=boldsymbol{a}+boldsymbol{b}

The order that you add the vectors in doesn’t matter, the resultant vector is always the same.

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Subtracting Vectors

Subtracting a vector is the same as adding a negative vector.

If aboldsymbol{a} is a vector, then aboldsymbol{-a} is a vector of the same size, in the opposite direction.

Therefore, subtracting a vector is the same as adding the negative vector.

E.g. ab=a+(b)boldsymbol{a-b}=boldsymbol{a+(-b)}

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Multiplying Vectors by Scalars

When you multiply a vector by a scalar (numeric value), this changes the length of the vector, but has no affect on the direction.

If a vector is multiplied by a non-zero scalar, the new vector is always parallel to the original vector, e.g. aboldsymbol{a} is parallel to 3a3boldsymbol{a}

To show two vectors are parallel you need to show that they are scalar multiples of each other.

Example: AB=xoverrightarrow{AB}=boldsymbol{x}, BC=yoverrightarrow{BC}=boldsymbol{y}

The point MM lies on the line ABAB and divides ABoverrightarrow{AB} in the ratio 3:23:2 and NN lies on the line BCBC and divides BCoverrightarrow{BC} in the ratio 2:32:3

Show that MNoverrightarrow{MN} is parallel to ACoverrightarrow{AC}

 

AC=x+yoverrightarrow{AC}=boldsymbol{x}+boldsymbol{y}

MM divides ABoverrightarrow{AB} in the ratio 3:23:2 so MM is 35dfrac{3}{5} of the way along ABoverrightarrow{AB}. Therefore, AM=35xoverrightarrow{AM}=dfrac{3}{5}boldsymbol{x}, so MB=25xoverrightarrow{MB}=dfrac{2}{5}boldsymbol{x}.

Similarly, NN is 25dfrac{2}{5} of the way along BCoverrightarrow{BC}, so BN=25yoverrightarrow{BN}=dfrac{2}{5}boldsymbol{y}

Finally, MN=25x+25y=25(x+y)=25ACoverrightarrow{MN}=dfrac{2}{5}boldsymbol{x}+dfrac{2}{5}boldsymbol{y}=dfrac{2}{5}(boldsymbol{x}+boldsymbol{y})=dfrac{2}{5}overrightarrow{AC}, which shows MNoverrightarrow{MN} is parallel to ACoverrightarrow{AC}

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Vector Basics Example Questions

Question 1: Find the resultant vector rboldsymbol{r} of the two single vectors aboldsymbol{a} and 2b2boldsymbol{b} using the diagram below.

[1 mark]

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Going from the start of aboldsymbol{a} to the end of 2b2boldsymbol{b}, r=a+2bboldsymbol{r}=boldsymbol{a}+2boldsymbol{b}

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Question 2: Using the diagram below, write down the following vectors in terms of a,bboldsymbol{a},boldsymbol{b} and cboldsymbol{c}

a) XYoverrightarrow{XY}

b) XZoverrightarrow{XZ}

c) ZYoverrightarrow{ZY}

[3 marks]

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XY=c+aoverrightarrow{XY}=-boldsymbol{c}+boldsymbol{a}

XZ=c+boverrightarrow{XZ}=-boldsymbol{c}+boldsymbol{b}

ZY=b+aoverrightarrow{ZY}=-boldsymbol{b}+boldsymbol{a}

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Question 3: XY=poverrightarrow{XY}=boldsymbol{p}, XZ=qoverrightarrow{XZ}=boldsymbol{q}

AA divides XYoverrightarrow{XY} in the ratio 5:15:1 and BB divides XZoverrightarrow{XZ} in the ratio 5:15:1

Show that YZoverrightarrow{YZ} is parallel to ABoverrightarrow{AB}

[3 marks]

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YZ=YX+XZ=p+qoverrightarrow{YZ}=overrightarrow{YX}+overrightarrow{XZ}=-boldsymbol{p}+boldsymbol{q}

AA divides XYoverrightarrow{XY} in the ratio 5:15:1 so AA is 56dfrac{5}{6} of the way along XYoverrightarrow{XY}. Therefore, AX=56poverrightarrow{AX}=-dfrac{5}{6}boldsymbol{p}. [1.2em]

Similarly, BB is 56dfrac{5}{6} of the way along XZoverrightarrow{XZ}, so XB=56qoverrightarrow{XB}=dfrac{5}{6}boldsymbol{q}[1.2em]

Finally, AB=56p+56q=56(p+q)=56YZoverrightarrow{AB}=-dfrac{5}{6}boldsymbol{p}+dfrac{5}{6}boldsymbol{q}=dfrac{5}{6}(-boldsymbol{p}+boldsymbol{q})=dfrac{5}{6}overrightarrow{YZ}, which shows ABoverrightarrow{AB} is parallel to YZoverrightarrow{YZ}

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Additional Resources

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Exam Tips Cheat Sheet

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Formula Booklet

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Specification Points Covered

J1 – Use vectors in two dimensions
J2 – Calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form