Mathematics 9709 · Pure Mathematics 3

Unit vectors

A unit vector points the same way as $\mathbf{a}$ but has length exactly 1. It keeps the direction and throws the size away.
To get it, divide every component by the length. It is written $\hat{\mathbf{a}}$ (say “a-hat”).

$\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}$ : find the length first, then divide the whole vector by that one number. Same direction, length now 1.

Quick check: square and add the components of a unit vector and you must get $1$ . If you don't, you divided by the wrong number. It takes a few seconds and saves the mark.

Divide by the length: â = a / |a|

$\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}=\dfrac{1}{|\mathbf{a}|}\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$

First find the length $|\mathbf{a}|$ by Pythagoras, then shrink the whole vector by that number.

â lies along a but is shrunk to length 1: same direction, size 1.

Why dividing by the length works

Multiplying a vector by a number $k$ multiplies its length by $k$ but keeps its direction (when $k>0$ ). So if you multiply $\mathbf{a}$ by exactly $\tfrac{1}{|\mathbf{a}|}$ , the new length is $\tfrac{1}{|\mathbf{a}|}\times|\mathbf{a}|=1$ . You get a vector of length 1 pointing the same way. That is all a unit vector is.

Worked example

Unit vector in the direction of $\begin{pmatrix}5\\2\end{pmatrix}$

Length first: $|\mathbf{a}|=\sqrt{5^2+2^2}=\sqrt{29}$ .
Divide each component by $\sqrt{29}$ , keeping the surd exact.

Answer

\hat{\mathbf{a}}=\dfrac{1}{\sqrt{29}}\begin{pmatrix}5\\2\end{pmatrix}=\begin{pmatrix}5/\sqrt{29}\\[2pt]2/\sqrt{29}\end{pmatrix}

Where the marks go

Common mistake

Divide by the whole length, not by one component. And keep the

\tfrac{1}{|\mathbf{a}|}

as an exact surd unless asked for a decimal. So leave it as

\tfrac{1}{\sqrt{6}}(-\mathbf{i}+\mathbf{j}+2\mathbf{k})

; that is what the mark scheme wants. (9709/33 Oct/Nov 2023 Q11)

Now you try

The line $l$ has equation $\mathbf{r}=\mathbf{i}-2\mathbf{j}-3\mathbf{k}+\lambda(-\mathbf{i}+\mathbf{j}+2\mathbf{k})$ . Find a unit vector pointing along $l$ . (9709/33 Oct/Nov 2023 Q11)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2023 Q11(a))

The direction is the bracket next to $\lambda$ : $\mathbf{d}=(-1,1,2)$ , with $|\mathbf{d}|=\sqrt{(-1)^2+1^2+2^2}=\sqrt{6}$ :

Answer

\hat{\mathbf{d}}=\dfrac{1}{\sqrt6}(-\mathbf{i}+\mathbf{j}+2\mathbf{k})

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Unit vectors

A unit vector points the same way as $\mathbf{a}$ but has length exactly 1. It keeps the direction and throws the size away.
To get it, divide every component by the length. It is written $\hat{\mathbf{a}}$ (say “a-hat”).

$\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}$ : find the length first, then divide the whole vector by that one number. Same direction, length now 1.

Quick check: square and add the components of a unit vector and you must get $1$ . If you don't, you divided by the wrong number. It takes a few seconds and saves the mark.

Divide by the length: â = a / |a|

$\hat{\mathbf{a}}=\dfrac{\mathbf{a}}{|\mathbf{a}|}=\dfrac{1}{|\mathbf{a}|}\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$

First find the length $|\mathbf{a}|$ by Pythagoras, then shrink the whole vector by that number.

â lies along a but is shrunk to length 1: same direction, size 1.

Why dividing by the length works

Worked example

Unit vector in the direction of $\begin{pmatrix}5\\2\end{pmatrix}$

Length first: $|\mathbf{a}|=\sqrt{5^2+2^2}=\sqrt{29}$ .
Divide each component by $\sqrt{29}$ , keeping the surd exact.

Answer

\hat{\mathbf{a}}=\dfrac{1}{\sqrt{29}}\begin{pmatrix}5\\2\end{pmatrix}=\begin{pmatrix}5/\sqrt{29}\\[2pt]2/\sqrt{29}\end{pmatrix}

Where the marks go

Common mistake

Divide by the whole length, not by one component. And keep the

\tfrac{1}{|\mathbf{a}|}

as an exact surd unless asked for a decimal. So leave it as

\tfrac{1}{\sqrt{6}}(-\mathbf{i}+\mathbf{j}+2\mathbf{k})

; that is what the mark scheme wants. (9709/33 Oct/Nov 2023 Q11)

Now you try

The line $l$ has equation $\mathbf{r}=\mathbf{i}-2\mathbf{j}-3\mathbf{k}+\lambda(-\mathbf{i}+\mathbf{j}+2\mathbf{k})$ . Find a unit vector pointing along $l$ . (9709/33 Oct/Nov 2023 Q11)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2023 Q11(a))

The direction is the bracket next to $\lambda$ : $\mathbf{d}=(-1,1,2)$ , with $|\mathbf{d}|=\sqrt{(-1)^2+1^2+2^2}=\sqrt{6}$ :

Answer

\hat{\mathbf{d}}=\dfrac{1}{\sqrt6}(-\mathbf{i}+\mathbf{j}+2\mathbf{k})