Mathematics 9709 · Pure Mathematics 3

Foot of perpendicular & shortest distance

The shortest way from a point to a line is the perpendicular drop. Where it lands on the line is the foot of the perpendicular.
The whole method rests on one idea. Take a general point on the line. The foot is the spot where the join from your point meets the line at a right angle (their dot product is zero).

The foot. Write $P=\mathbf{a}+t\mathbf{d}$ , build $\overrightarrow{AP}$ , set $\overrightarrow{AP}\cdot\mathbf{d}=0$ and solve for $t$ . That one line fixes the foot. Then put $t$ back into $P$ .

The distance. Once you have the foot, the shortest distance is just the length of the drop, $|\overrightarrow{AP}|$ . The same method also gives the reflection of $A$ in the line: the foot is the midpoint, so the image is $2\overrightarrow{OP}-\mathbf{a}$ .

General point, dot product = 0, solve for t

Write a general point on the line: $P=\mathbf{a}+t\mathbf{d}$ (its coordinates now carry $t$ ).
Form the join from your fixed point $A$ : $\overrightarrow{AP}=P-\mathbf{a}$ .
Make it perpendicular to the direction: $\overrightarrow{AP}\cdot\mathbf{d}=0$ , then solve for $t$ .
Put that $t$ back into $P$ for the foot; the shortest distance is $|\overrightarrow{AP}|$ .

P left–rightP up–down

shortest distance PF = 2.32·foot F = (6.9, 3.8)·the angle at F stays 90°

Move P and watch the join PF: it is shortest when it meets the line at a right angle. That right angle is the $\overrightarrow{AP}\cdot\mathbf{d}=0$ the method rests on.

Why perpendicular is the shortest of all

Take any other point $Q$ on the line. Then $A$ , the foot $P$ and $Q$ make a right-angled triangle with the right angle at $P$ . Here $AQ$ is the longest side (the hypotenuse). The hypotenuse is always the longest side, so $AP<AQ$ for every other $Q$ . The straight-down drop beats every slanted one. That is why “shortest distance” and “perpendicular” are the same problem.

Worked example

Real question: foot of perpendicular from $A=\mathbf{i}+5\mathbf{j}+6\mathbf{k}$ to $\mathbf{r}=4\mathbf{i}+\mathbf{k}+t(-\mathbf{i}+2\mathbf{j}+3\mathbf{k})$

9709/33 M/J 2022 Q9(b). This one method also handles the acute-angle part (a) and the reflection part (c).

General point on the line: $P=(4-t,\ 2t,\ 1+3t)$ .
Join from A: $\overrightarrow{AP}=P-A=(3-t,\ 2t-5,\ 3t-5)$ .
Set $\overrightarrow{AP}\cdot\mathbf{d}=0$ with $\mathbf{d}=(-1,2,3)$ : $-(3-t)+2(2t-5)+3(3t-5)=0$ .
Expand: $-3+t+4t-10+9t-15=0\Rightarrow14t-28=0\Rightarrow t=2$ .
Put $t=2$ back into $P$ .

Answer

\overrightarrow{OP}=2\mathbf{i}+4\mathbf{j}+7\mathbf{k}

Want the shortest distance too? It is $|\overrightarrow{AP}|=|(1,-1,1)|=\sqrt3$ , the length of that perpendicular drop.

Where the marks go

Common mistake

Dot

\overrightarrow{AP}

with the direction

\mathbf{d}

, not with

\mathbf{a}

\overrightarrow{OP}

. The right angle is between the join and the way the line points.

Common mistake

The foot is the position vector

\mathbf{a}+t\mathbf{d}

, so put $t$ back in. The value of

t

on its own is not the answer. And the distance is

|\overrightarrow{AP}|

, not the foot itself, so read which one the question wants. (9709/32 May/June 2023 Q11)

Now you try

The point $A$ has position vector $\mathbf{i}+2\mathbf{j}-2\mathbf{k}$ and the line $l$ is $\mathbf{r}=\mathbf{i}-\mathbf{j}+3\mathbf{k}+t(2\mathbf{i}-3\mathbf{j}+4\mathbf{k})$ . Find the position vector of the foot of the perpendicular from $A$ to $l$ . (9709/32 May/June 2023 Q11)

Your turn— tap to reveal the worked answer (9709/32 May/June 2023 Q11(b))

General point $P=(1+2t,\ -1-3t,\ 3+4t)$ , so the join is $\overrightarrow{AP}=(2t,\ -3-3t,\ 5+4t)$ .

Set the dot to zero: $\overrightarrow{AP}\cdot(2,-3,4)=4t+9+9t+20+16t=29t+29=0\ \Rightarrow\ t=-1$ . Put it back:

Answer

\overrightarrow{OP}=-\mathbf{i}+2\mathbf{j}-\mathbf{k}

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Foot of perpendicular & shortest distance

The shortest way from a point to a line is the perpendicular drop. Where it lands on the line is the foot of the perpendicular.
The whole method rests on one idea. Take a general point on the line. The foot is the spot where the join from your point meets the line at a right angle (their dot product is zero).

General point, dot product = 0, solve for t

Write a general point on the line: $P=\mathbf{a}+t\mathbf{d}$ (its coordinates now carry $t$ ).
Form the join from your fixed point $A$ : $\overrightarrow{AP}=P-\mathbf{a}$ .
Make it perpendicular to the direction: $\overrightarrow{AP}\cdot\mathbf{d}=0$ , then solve for $t$ .
Put that $t$ back into $P$ for the foot; the shortest distance is $|\overrightarrow{AP}|$ .

P left–rightP up–down

shortest distance PF = 2.32·foot F = (6.9, 3.8)·the angle at F stays 90°

Move P and watch the join PF: it is shortest when it meets the line at a right angle. That right angle is the $\overrightarrow{AP}\cdot\mathbf{d}=0$ the method rests on.

Why perpendicular is the shortest of all

Worked example

Real question: foot of perpendicular from $A=\mathbf{i}+5\mathbf{j}+6\mathbf{k}$ to $\mathbf{r}=4\mathbf{i}+\mathbf{k}+t(-\mathbf{i}+2\mathbf{j}+3\mathbf{k})$

9709/33 M/J 2022 Q9(b). This one method also handles the acute-angle part (a) and the reflection part (c).

General point on the line: $P=(4-t,\ 2t,\ 1+3t)$ .
Join from A: $\overrightarrow{AP}=P-A=(3-t,\ 2t-5,\ 3t-5)$ .
Set $\overrightarrow{AP}\cdot\mathbf{d}=0$ with $\mathbf{d}=(-1,2,3)$ : $-(3-t)+2(2t-5)+3(3t-5)=0$ .
Expand: $-3+t+4t-10+9t-15=0\Rightarrow14t-28=0\Rightarrow t=2$ .
Put $t=2$ back into $P$ .

Answer

\overrightarrow{OP}=2\mathbf{i}+4\mathbf{j}+7\mathbf{k}

Want the shortest distance too? It is $|\overrightarrow{AP}|=|(1,-1,1)|=\sqrt3$ , the length of that perpendicular drop.

Where the marks go

Common mistake

Dot

\overrightarrow{AP}

with the direction

\mathbf{d}

, not with

\mathbf{a}

\overrightarrow{OP}

. The right angle is between the join and the way the line points.

Common mistake

The foot is the position vector

\mathbf{a}+t\mathbf{d}

, so put $t$ back in. The value of

t

on its own is not the answer. And the distance is

|\overrightarrow{AP}|

, not the foot itself, so read which one the question wants. (9709/32 May/June 2023 Q11)

Now you try

Your turn— tap to reveal the worked answer (9709/32 May/June 2023 Q11(b))

General point $P=(1+2t,\ -1-3t,\ 3+4t)$ , so the join is $\overrightarrow{AP}=(2t,\ -3-3t,\ 5+4t)$ .

Set the dot to zero: $\overrightarrow{AP}\cdot(2,-3,4)=4t+9+9t+20+16t=29t+29=0\ \Rightarrow\ t=-1$ . Put it back:

Answer

\overrightarrow{OP}=-\mathbf{i}+2\mathbf{j}-\mathbf{k}