In the triangle shown below, one vertex is origin 𝑂, and the two other vertices are $𝐴$ and \(𝐵

Question

In the triangle shown below, one vertex is origin 𝑂, and the two other vertices are $𝐴$ and $𝐵$ where $\overrightarrow{𝑂𝐴} = \mathbf{a}, \overrightarrow{𝑂𝐵} = \mathbf{b}$. A median of a triangle is a line segment joining a vertex $(O,A$ or $B)$to the midpoint of the opposite side (e.g. $AD$ is a median of triangle $OAB$ from vertex $A$). It is given that $F$ is the point of intersection between the medians of triangle $OAB$ from vertices $A$ and $B$. By finding $\overrightarrow{AD}$ and $\overrightarrow{BE}$, show that $\overrightarrow{OF}=\frac{1}{3}\left( \mathbf{a}+\mathbf{b} \right)$.[3] Prove that $F$ also lies on $OC$, the median of triangle $OAB$ from vertex $O$.[2] With reference to the origin $O$, the points $N$, $P$ and $Q$ have position vectors $\mathbf{n}$,$\mathbf{p}$ and $\mathbf{q}$ respectively, and $\prod{{}}$ has an equation $\mathbf{r}\cdot \mathbf{n}=0$. It is known that $\mathbf{p}\cdot \mathbf{n}=\mathbf{q}\cdot \mathbf{n}\ne 0$. Show that $\overrightarrow{PQ}$ is perpendicular to $\mathbf{n}$. Hence describe the geometrical relationship between the line $PQ$ and the plane $\prod{{}}$.[3] Find a vector equation of the plane that contains points $P$ and $Q$, and is perpendicular to plane $\prod{{}}$, leaving your answer in terms of $\mathbf{n}$,$\mathbf{p}$ and $\mathbf{q}$.[2]

Accepted Answer

(a)(i) OF = 1/3(a+b); (a)(ii) Since ${\overrightarrow{OF} = k\overrightarrow{OC}}$ for some scalar ${k}$ where ${0 < k < 1}$, point ${F}$ lies on ${OC}$. (b)(i) PQ⊥n, PQ∥Π; (b)(ii) r=p+λ(q-p)+μn or r·[(q-p)×n]=p·[(q-p)×n]

2020 CJC P2 Q5

2020 CJC P2 Q5

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