DHS 2025 H2 Math Prelim Paper 2 — Vectors (worked solution)

(a) $l_{OD}:\; \mathbf{r} = \lambda\bigl(k\mathbf{b} + (1-k)\mathbf{a}\bigr), | \lambda \in \mathbb{R}$ | (b) $\lambda = \dfrac{5}{2}$, | $k = \dfrac{2}{5}$ | (c) $\theta = 104.5^{\circ}$ (1 d.p.) | (d) $|\mathbf{a}\cdot\mathbf{b}|$ is the

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Vectors: Pos. vectors; collinearity; angle — DHS 2025 H2 Math Prelim Paper 2

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Dunman High School (DHS)2025 PrelimPaper 2●●○ Standard11 marks

What this question tests

Pos. vectors; collinearity; angle.

Question

Figure for this question

Relative to the origin $O$, the points $A$, $B$, and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $3\mathbf{a}+\mathbf{b}$ respectively. The point $D$ lies on $AB$ such that $AD = kAB$, where $0 < k < 1$.

Find a vector equation of the line $OD$ in terms of $k$, $\mathbf{a}$ and $\mathbf{b}$.
The point $E$ is the midpoint of $BC$. Find the value of $k$ if $O$, $D$ and $E$ are collinear.

It is given that $|\mathbf{a}| = 1$, $|\mathbf{b}| = 2$ and $|3\mathbf{a}-2\mathbf{b}| = \sqrt{31}$.

By considering the scalar product $(3\mathbf{a}-2\mathbf{b})\cdot(3\mathbf{a}-2\mathbf{b})$, find the numerical value of $\mathbf{a}\cdot\mathbf{b}$ and hence determine the angle between $\mathbf{a}$ and $\mathbf{b}$.
Give a geometrical interpretation of $|\mathbf{a}\cdot\mathbf{b}|$.

Show full worked solution▾

(a)

Let $\overrightarrow{AD} = k\overrightarrow{AB}$.

\[\overrightarrow{OD} = \frac{k\mathbf{b} + (1-k)\mathbf{a}}{k+(1-k)} = k\mathbf{b} + (1-k)\mathbf{a}\]

The question asks for a vector equation of line $OD$. Line $OD \neq \overrightarrow{OD}$. The equation of line $OD$ is $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ where $\mathbf{a}$ is the position vector of a fixed point on the line and $\mathbf{b}$ is the direction vector.

(b)

$\overrightarrow{OC} = 3\mathbf{a} + \mathbf{b}$, $\overrightarrow{OB} = \mathbf{b}$.

\[\overrightarrow{OE} = \tfrac{1}{2}\!\left(\overrightarrow{OC} + \overrightarrow{OB}\right) = \tfrac{1}{2}(3\mathbf{a} + \mathbf{b} + \mathbf{b}) = \tfrac{1}{2}(3\mathbf{a} + 2\mathbf{b})\]

Since $O$, $D$ and $E$ are collinear, $E$ lies on $l_{OD}$: \[\overrightarrow{OE} = \lambda\bigl(k\mathbf{b} + (1-k)\mathbf{a}\bigr)\] \[\tfrac{3}{2}\mathbf{a} + \mathbf{b} = \lambda(1-k)\mathbf{a} + \lambda k\mathbf{b}\]

Since $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel: \[\begin{aligned} \lambda k &= 1 \quad \cdots (1)\\ \lambda(1-k) &= \tfrac{3}{2} \quad \cdots (2) \end{aligned}\] Substituting (1) into (2): $\lambda - 1 = \tfrac{3}{2} \Rightarrow \lambda = \tfrac{5}{2}$.

(c)

$|\mathbf{a}| = 1$, $|\mathbf{b}| = 2$, $|3\mathbf{a} - 2\mathbf{b}| = \sqrt{31}$.

\[\begin{aligned} |3\mathbf{a} - 2\mathbf{b}|^2 &= 31\\ (3\mathbf{a} - 2\mathbf{b})\cdot(3\mathbf{a} - 2\mathbf{b}) &= 31\\ 9|\mathbf{a}|^2 - 12\,\mathbf{a}\cdot\mathbf{b} + 4|\mathbf{b}|^2 &= 31\\ 9(1)^2 - 12\,\mathbf{a}\cdot\mathbf{b} + 4(2)^2 &= 31\\ 25 - 12\,\mathbf{a}\cdot\mathbf{b} &= 31\\ \mathbf{a}\cdot\mathbf{b} &= -\tfrac{1}{2} \end{aligned}\]

Let $\theta$ be the angle between $\mathbf{a}$ and $\mathbf{b}$: \[\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{-\frac{1}{2}}{1 \times 2} = -\frac{1}{4}\]

$(3\mathbf{a}-2\mathbf{b})\cdot(3\mathbf{a}-2\mathbf{b}) = |3\mathbf{a}-2\mathbf{b}|^2 \neq (3\mathbf{a}-2\mathbf{b})^2$. The question does not ask for the acute angle, so modulus for $\mathbf{a}\cdot\mathbf{b}$ should not be applied.

(d)

$|\mathbf{a}\cdot\mathbf{b}|$ is the length of projection of $\mathbf{b}$ onto $\mathbf{a}$.

Answer: (a) $l_{OD}:\; \mathbf{r} = \lambda\bigl(k\mathbf{b} + (1-k)\mathbf{a}\bigr),\quad \lambda \in \mathbb{R}$ (b) $\lambda = \dfrac{5}{2}$, $k = \dfrac{2}{5}$ (c) $\theta = 104.5^{\circ}$ (1 d.p.) (d) $|\mathbf{a}\cdot\mathbf{b}|$ is the length of projection of $\mathbf{b}$ onto $\mathbf{a}$.

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