Scalar product and vector angles | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Connect components, angle and perpendicularity

Both vectors leave the angle vertex

uvθ

u · v = |u||v| cos θ

For an angle at B, build BA and BC so both arrows start at B.

\mathbf u\cdot\mathbf v=u_1v_1+u_2v_2+u_3v_3

\mathbf u\cdot\mathbf v=|\mathbf u||\mathbf v|\cos\theta

The vector angle $\theta$ is measured between 0° and 180°.
A positive dot product gives an acute vector angle.
A negative dot product gives an obtuse vector angle.
For non-zero vectors, $\mathbf u\cdot\mathbf v=0$ means perpendicular.

To find an angle, first calculate the exact cosine ratio, then use the $\cos^{-1}$ key in degree mode. Unless an exact answer is requested, give the final angle to 3 significant figures.

For angle ABC, make both arrows leave B

Worked example

Current-paper angle at B

For $A=(2,1,-3)$ , $B=(0,4,1)$ and $C=(-3,-2,2)$ :

\overrightarrow{BA}=(2,-3,-4)

\overrightarrow{BC}=(-3,-6,1)

\overrightarrow{BA}\cdot\overrightarrow{BC}=8

\cos\theta=\frac8{\sqrt{29}\sqrt{46}}

Answer

\theta=77.3^\circ

(9709/32/O/N/24 Q9(c))

Common mistake

\overrightarrow{AB}

and

\overrightarrow{BC}

do not both leave B. Use

\overrightarrow{BA}

and

\overrightarrow{BC}

In a 3D object, build both vectors from one vertex

Add the edge moves from O

OPR4i3j2k

OP = 4i + 3j + 2k

OR = 4i + 3j

Perspective sketch: use the labelled vectors, not visual scale.

In a cuboid, reach a vertex by adding the edge moves; then use those vectors in the scalar product.

The perspective sketch is not a measuring tool. Add the labelled edge vectors to reach each point, then use the scalar product.

Your turnIndependent check [4 marks]

In the cuboid shown, $\overrightarrow{OP}=(4,3,2)$ and $\overrightarrow{OR}=(4,3,0)$ . Find angle POR.

Show worked answer

\overrightarrow{OP}\cdot\overrightarrow{OR}=25

|OP|=\sqrt{29},\qquad |OR|=5

\cos\theta=\frac{25}{5\sqrt{29}}=\frac5{\sqrt{29}}

Answer

\theta=21.8^\circ

(9709 syllabus §3.7; Coursebook, Ch. 9 Exercise 9C)

Connect components, angle and perpendicularity

Both vectors leave the angle vertex

uvθ

u · v = |u||v| cos θ

For an angle at B, build BA and BC so both arrows start at B.

\mathbf u\cdot\mathbf v=u_1v_1+u_2v_2+u_3v_3

\mathbf u\cdot\mathbf v=|\mathbf u||\mathbf v|\cos\theta

The vector angle $\theta$ is measured between 0° and 180°.
A positive dot product gives an acute vector angle.
A negative dot product gives an obtuse vector angle.
For non-zero vectors, $\mathbf u\cdot\mathbf v=0$ means perpendicular.

To find an angle, first calculate the exact cosine ratio, then use the $\cos^{-1}$ key in degree mode. Unless an exact answer is requested, give the final angle to 3 significant figures.

For angle ABC, make both arrows leave B

Worked example

Current-paper angle at B

For $A=(2,1,-3)$ , $B=(0,4,1)$ and $C=(-3,-2,2)$ :

\overrightarrow{BA}=(2,-3,-4)

\overrightarrow{BC}=(-3,-6,1)

\overrightarrow{BA}\cdot\overrightarrow{BC}=8

\cos\theta=\frac8{\sqrt{29}\sqrt{46}}

Answer

\theta=77.3^\circ

(9709/32/O/N/24 Q9(c))

Common mistake

\overrightarrow{AB}

and

\overrightarrow{BC}

do not both leave B. Use

\overrightarrow{BA}

and

\overrightarrow{BC}

In a 3D object, build both vectors from one vertex

Add the edge moves from O

OPR4i3j2k

OP = 4i + 3j + 2k

OR = 4i + 3j

Perspective sketch: use the labelled vectors, not visual scale.

In a cuboid, reach a vertex by adding the edge moves; then use those vectors in the scalar product.

The perspective sketch is not a measuring tool. Add the labelled edge vectors to reach each point, then use the scalar product.

Your turnIndependent check [4 marks]

In the cuboid shown, $\overrightarrow{OP}=(4,3,2)$ and $\overrightarrow{OR}=(4,3,0)$ . Find angle POR.

Show worked answer

\overrightarrow{OP}\cdot\overrightarrow{OR}=25

|OP|=\sqrt{29},\qquad |OR|=5

\cos\theta=\frac{25}{5\sqrt{29}}=\frac5{\sqrt{29}}

Answer

\theta=21.8^\circ

(9709 syllabus §3.7; Coursebook, Ch. 9 Exercise 9C)