Orthogonality and an angle maximisation · source reference

Question

A,B,C have position vectors a,b,c relative to O. $\overrightarrow{AC}=4\overrightarrow{CB}$ and $|\mathbf a+\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2$.

(i) Consider $(\mathbf a+\mathbf b)\cdot(\mathbf a+\mathbf b)$ to show a and b are perpendicular. [2]

(ii) Find the length of projection of c on a in terms of $|\mathbf a|$. [3]

(iii) F is the perpendicular foot from C to OA; f is its position vector. State the geometrical meaning of $|\mathbf c\times\mathbf f|$. [1]

(iv) Two points X,Y move along line segments OA,AB respectively such that

$$\overrightarrow{OX}=(\cos3t)\mathbf i+(\sin3t)\mathbf j+\tfrac12\mathbf k,$$ $$\overrightarrow{OY}=(\sin t)\mathbf i+(\cos t)\mathbf j-2\mathbf k,$$

where t is real, $0\leq t\leq2\pi$. Express their scalar product as $p\sin(qt)+r$, with real p,q,r to determine, and find the greatest $\angle XOY$. [5]

[2017/ACJC/Prelim/I/9]

Source working

(i)

$$(\mathbf a+\mathbf b)\cdot(\mathbf a+\mathbf b)=\mathbf a\cdot\mathbf a+2\mathbf a\cdot\mathbf b+\mathbf b\cdot\mathbf b.$$

This equals $|\mathbf a+\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2$, so

$$|\mathbf a|^2+2\mathbf a\cdot\mathbf b+|\mathbf b|^2=|\mathbf a|^2+|\mathbf b|^2\Rightarrow2\mathbf a\cdot\mathbf b=0\Rightarrow\mathbf a\cdot\mathbf b=0\Rightarrow\mathbf a\perp\mathbf b.$$

(ii) Ratio theorem:

$$\overrightarrow{OC}=\frac{4\mathbf b+\mathbf a}5=\tfrac15\mathbf a+\tfrac45\mathbf b.$$

Projection length:

$$\begin{aligned}OF&=\frac{|\overrightarrow{OC}\cdot\overrightarrow{OA}|}{|OA|}\\&=\frac{|(\tfrac15\mathbf a+\tfrac45\mathbf b)\cdot\mathbf a|}{|\mathbf a|}\\&=\frac{|\tfrac15\mathbf a\cdot\mathbf a+\tfrac45\mathbf b\cdot\mathbf a|}{|\mathbf a|}\\&=\frac{|\tfrac15|\mathbf a|^2+\tfrac45\mathbf b\cdot\mathbf a|}{|\mathbf a|}=\tfrac15|\mathbf a|.\end{aligned}$$

(iii) $|\mathbf c\times\mathbf f|$ is twice the area of triangle COF.

(iv)

$$\begin{aligned}\overrightarrow{OX}\cdot\overrightarrow{OY}&=(\cos3t,\sin3t,1/2)\cdot(\sin t,\cos t,-2)\\&=\cos3t\sin t+\sin3t\cos t-1=\sin4t-1.\end{aligned}$$ $$\begin{aligned}\cos\angle XOY&=\frac{\sin4t-1}{\sqrt{\cos^23t+\sin^23t+1/4}\sqrt{\sin^2t+\cos^2t+4}}\\&=\frac{\sin4t-1}{\sqrt{5/4}\sqrt5}=\tfrac25(\sin4t-1).\end{aligned}$$

Maximum angle occurs when its cosine is most negative, when $\sin4t=-1$. Cosine decreases as θ increases from 0 to π: larger angle, smaller cosine. Then $\cos\angle XOY=\tfrac25(-1-1)=-4/5$, so $\angle XOY=\cos^{-1}(-4/5)=143.1^\circ$.

Qualification

The printed assertion that X,Y move along fixed line segments contradicts the supplied trigonometric positions: those describe horizontal circles. The stated angle result follows the formulas; the line-segment narrative cannot simultaneously hold for the entire t interval.

Native diagrams

The projection triangle labels A,C,B collinearly, $AC:CB=4:1$, O, perpendicular F on OA, red OF and dashed CF. The cosine graph labels $y=\cos\theta$, θ-axis, y-axis, 0, $\pi/2,\pi,1,-1$, with projection guides at the minimum $(\pi,-1)$.

Printed answer key

(ii) $|\mathbf a|/5$; (iii) twice the area of triangle COF; (iv) $\sin4t-1$, $143.1^\circ$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 2; solution document pages 9,10. Original page numbers are retained as text metadata.