Foot of a perpendicular from a point | CAIE 9709 Pure Mathematics 3 Notes

Chapters

Pure Mathematics 3

Mathematics9709

Put H on the line, then force a right angle

The foot makes a right angle

QH = a + tdd

QH · d = 0

Put a general point H on the line, then enforce QH · d = 0.

Write a general line point $\mathbf h=\mathbf a+t\mathbf d$ .
Form $\overrightarrow{QH}=\mathbf h-\mathbf q$ .
Solve $(\mathbf h-\mathbf q)\cdot\mathbf d=0$ for $t$ .
Substitute back to get H. If distance is required, calculate $|\overrightarrow{QH}|$ .

Here $\mathbf h$ and $\mathbf q$ are the position vectors of H and Q. Their difference is the directed join from Q to H.

This perpendicular is the shortest route from Q to the line. Any other line point makes a right-angled triangle in which the sloping route from Q is the hypotenuse, so it must be longer than QH.

Examiner note

The shortest distance between two skew lines and a common perpendicular to skew lines are not required. The vector product is also outside this syllabus.

The dot product creates one equation

Worked example

Foot and point-to-line distance

Let $Q=(1,2,3)$ and let the line be:

\mathbf r=(0,1,-1)+t(2,1,2)

\mathbf h=(2t,1+t,-1+2t)

\overrightarrow{QH}=\begin{pmatrix}2t-1\\t-1\\2t-4\end{pmatrix}

(2t-1,t-1,2t-4)

\cdot(2,1,2)=0

9t-11=0,\qquad t=\frac{11}{9}

Answer

H=\left(\frac{22}{9},\frac{20}{9},\frac{13}{9}\right)

QH=\frac{\sqrt{41}}3

(Pure Mathematics 2 & 3 Coursebook, Ch. 9 Exercise 9D and review)

Check both membership and perpendicularity

Your turnIndependent check [5 marks]

Find the foot from $Q=(2,0,1)$ to the line $\mathbf r=(1,1,0)+t(1,-1,2)$ , and hence find the distance from Q to the line.

Show worked answer

\mathbf h=(1+t,1-t,2t)

\overrightarrow{QH}=\begin{pmatrix}t-1\\1-t\\2t-1\end{pmatrix}

\overrightarrow{QH}\cdot(1,-1,2)=0

6t-4=0

Answer

t=\frac23

H=\left(\frac53,\frac13,\frac43\right)

QH=\frac{\sqrt3}{3}

Put H on the line, then force a right angle

The foot makes a right angle

QH = a + tdd

QH · d = 0

Put a general point H on the line, then enforce QH · d = 0.

Write a general line point $\mathbf h=\mathbf a+t\mathbf d$ .
Form $\overrightarrow{QH}=\mathbf h-\mathbf q$ .
Solve $(\mathbf h-\mathbf q)\cdot\mathbf d=0$ for $t$ .
Substitute back to get H. If distance is required, calculate $|\overrightarrow{QH}|$ .

Here $\mathbf h$ and $\mathbf q$ are the position vectors of H and Q. Their difference is the directed join from Q to H.

This perpendicular is the shortest route from Q to the line. Any other line point makes a right-angled triangle in which the sloping route from Q is the hypotenuse, so it must be longer than QH.

Examiner note

The shortest distance between two skew lines and a common perpendicular to skew lines are not required. The vector product is also outside this syllabus.

The dot product creates one equation

Worked example

Foot and point-to-line distance

Let $Q=(1,2,3)$ and let the line be:

\mathbf r=(0,1,-1)+t(2,1,2)

\mathbf h=(2t,1+t,-1+2t)

\overrightarrow{QH}=\begin{pmatrix}2t-1\\t-1\\2t-4\end{pmatrix}

(2t-1,t-1,2t-4)

\cdot(2,1,2)=0

9t-11=0,\qquad t=\frac{11}{9}

Answer

H=\left(\frac{22}{9},\frac{20}{9},\frac{13}{9}\right)

QH=\frac{\sqrt{41}}3

(Pure Mathematics 2 & 3 Coursebook, Ch. 9 Exercise 9D and review)

Check both membership and perpendicularity

Your turnIndependent check [5 marks]

Find the foot from $Q=(2,0,1)$ to the line $\mathbf r=(1,1,0)+t(1,-1,2)$ , and hence find the distance from Q to the line.

Show worked answer

\mathbf h=(1+t,1-t,2t)

\overrightarrow{QH}=\begin{pmatrix}t-1\\1-t\\2t-1\end{pmatrix}

\overrightarrow{QH}\cdot(1,-1,2)=0

6t-4=0

Answer

t=\frac23

H=\left(\frac53,\frac13,\frac43\right)

QH=\frac{\sqrt3}{3}