Projection followed by line reflection · source reference

Question

Relative to origin O, A and B have position vectors $\mathbf a=3\mathbf i+\mathbf j+3\mathbf k$ and $\mathbf b=5\mathbf i-4\mathbf j+3\mathbf k$.

(i) Find the length of the projection of OA on OB. [2]

(ii) Hence or otherwise find the position vector of C on OB such that AC is perpendicular to OB. [2]

(iii) Find a vector equation of the reflection of line AB in line AC. [3]

[2010/ACJC/Prelim/I/9]

Source working

(i)

$$\frac{|\overrightarrow{OA}\cdot\overrightarrow{OB}|}{|\overrightarrow{OB}|}=\frac1{\sqrt{50}}|(3,1,3)\cdot(5,-4,3)|=\frac{20}{\sqrt{50}}=2\sqrt2\quad\text{or}\quad4/\sqrt2.$$

(ii) Line OB: $\mathbf r=(0,0,0)+\lambda(5,-4,3)$.

C on OB gives $\overrightarrow{OC}=\lambda(5,-4,3)$, $\lambda\in\mathbb R$.

$$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=\lambda(5,-4,3)-(3,1,3).$$

Since $AC\perp OB$, $\overrightarrow{AC}\cdot\overrightarrow{OB}=0$:

$$[\lambda(5,-4,3)-(3,1,3)]\cdot(5,-4,3)=0,$$ $$\lambda(25+16+9)-(15-4+9)=0,$$ $$\lambda=20/50=2/5,\qquad\overrightarrow{OC}=\tfrac25(5,-4,3).$$

Alternative: From (i), $|OC|=2\sqrt2$.

$$\overrightarrow{OC}=2\sqrt2\frac{\overrightarrow{OB}}{|\overrightarrow{OB}|}=2\sqrt2\frac1{\sqrt{50}}(5,-4,3)=\tfrac25(5,-4,3).$$

OC is in the same direction as OB since $\overrightarrow{OA}\cdot\overrightarrow{OB}>0$.

(iii) $\overrightarrow{OC}=\tfrac25\overrightarrow{OB}$, so $OC:CB=2:3$. Let B′ be the reflection of B in AC.

$$\overrightarrow{OB'}=-\tfrac15\overrightarrow{OB}=-\tfrac15(5,-4,3).$$ $$\overrightarrow{AB'}=\overrightarrow{OB'}-\overrightarrow{OA}=-\tfrac15(5,-4,3)-(3,1,3)=-\tfrac15(20,1,18).$$

The reflected line AB′ is

$$\mathbf r=(3,1,3)+\alpha(20,1,18),\qquad\alpha\in\mathbb R.$$

Teacher notes

Projection length equals $|\overrightarrow{OA}\cdot\overrightarrow{OB}|/|OB|=|OA||\cos\angle AOB|=OC$. C is the perpendicular foot from A to OB.

Alternative for (iii): C is the midpoint of B and B′. Ratio theorem about O:

$$\overrightarrow{OC}=\frac{\overrightarrow{OB}+\overrightarrow{OB'}}2,$$ $$\overrightarrow{OB'}=2\overrightarrow{OC}-\overrightarrow{OB}=\tfrac45\overrightarrow{OB}-\overrightarrow{OB}=-\tfrac15\overrightarrow{OB}=-\tfrac15(5,-4,3).$$

Native diagrams

Two projection diagrams place O,C,B collinearly with A above C and right angle ACB; the first highlights OC in red. The reflection diagram includes B′,O,C,B in order with consecutive lengths 1,2,3, A over C, original $l_{AB}$, reflected $l_{AB'}$, and a perpendicular at C.

Printed answer key

(i) $2\sqrt2$; (ii) $(2/5)(5,-4,3)$; (iii) $\mathbf r=(3,1,3)+\alpha(20,1,18)$, $\alpha\in\mathbb R$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 1; solution document pages 5,6. Original page numbers are retained as text metadata.