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

Question

Relative to the origin $O$, points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, where $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel vectors. The point $M$ is the midpoint of $OA$ and the point $N$ lies on $OB$ such that $ON:NB=1:2$. The point $C$ with position vector $\mathbf{c}$ is the point of intersection of lines $AN$ and $BM$. Show that $\mathbf{c}=\frac{2}{5}\mathbf{a}+\frac{1}{5}\mathbf{b}$. [3] Give the geometrical meaning of $\left| \mathbf{a}\cdot \widehat{\mathbf{c}} \right|$, where $\widehat{\mathbf{c}}$ is the unit vector in the direction of $\mathbf{c}$. [1] The points $O$, $A$ and $C$ lies on a circle where $OA$ is the diameter of the circle. It is further given that $\mathbf{a}\cdot \mathbf{b}=\frac{1}{2}{{\left| \mathbf{a} \right|}^{2}}$. Find the ratio of $\left| \mathbf{a} \right|:\left| \mathbf{c} \right|$. [3] Hence find the area of triangle $OAC$, giving your answer in terms of $\left| \mathbf{a} \right|$. [2]

Accepted Answer

(i) $\mathbf c=\frac25\mathbf a+\frac15\mathbf b$. (ii) The length of the orthogonal projection of $OA$ onto the line through $O$ parallel to $\mathbf c$. (iii) $\sqrt2:1$. (iv) $\frac14|\mathbf a|^2\text{ units}^2$.

2024 ACJC Promo Q6

2024 ACJC Promo Q6

Video Solution:

Solution:

Solution locked