Relative to the origin $O$, the points $A$, $B$, and $C$ have position vectors \(\mathbf{a}\

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 $\mathbf{b}$.[2] The point $E$ is the midpoint of $BC$. Find the value of $k$ if $O$, $D$ and $E$ are collinear.[4] 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 $\mathbf{b}$.[4] Give a geometrical interpretation of $|\mathbf{a} \cdot \mathbf{b}|$.[1]

Accepted Answer

(a) $\mathbf r=\lambda[(1-k)\mathbf a+k\mathbf b]$, $\lambda\in\mathbb R$ (b) $k=\dfrac25$ (c) $\mathbf a\cdot\mathbf b=-\dfrac12$, $\theta=104.5^\circ$ (d) length of the orthogonal projection of $\mathbf b$ in the direction of $\mathbf a$

2025 DHS P2 Q4

2025 DHS P2 Q4

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