The origin $O$ and the points $A$, $B$ and $C$ lies in the same plane, where \(\overrig

Question

The origin $O$ and the points $A$, $B$ and $C$ lies in the same plane, where $\overrightarrow{OA}=\mathbf{a}$, $\overrightarrow{OB}=\mathbf{b}$ and $\overrightarrow{OC}=\mathbf{c}$ (see diagram). Explain why $\mathbf{c}$ can be expressed as $\mathbf{c}=\lambda \mathbf{a}+\mu \mathbf{b}$, for constants $\lambda $ and $\mu $.[1] The point $N$ is on $AC$ such that $AN:NC=3:4$. Write down the position vector of $N$ in terms of $\mathbf{a}$ and $\mathbf{c}$.[1] It is given that the area of $ONC$ is equal to the area of triangle $OMC$, where $M$ is the mid-point $OB$. By finding the areas of these triangles in terms of $\mathbf{a}$ and $\mathbf{b}$, find $\lambda $ in terms of $\mu $ in the case where $\lambda $ and $\mu $ are both positive.[5]

Accepted Answer

(i) Equation of plane ${OABC}$ can be expressed as ${\mathbf{r} = \lambda\mathbf{a} + \mu\mathbf{b}, \lambda \in R, \mu \in R}$. Since ${C}$ lies on the plane ${OABC}$, ${\mathbf{c} = \lambda\mathbf{a} + \mu\mathbf{b}}$ for some for constants ${\lambda}$ and ${\mu}$. (ii) ${\overrightarrow{ON} = \frac{4\mathbf{a} + 3\mathbf{c}}{4 + 3} = \frac{1}{7}(4\mathbf{a} + 3\mathbf{c})}$ (iii) ${\lambda = \frac{8}{7}\mu}$

N2013 P1 Q6

N2013 P1 Q6

Video Solution:

Default solution