Let $A B C D$ be a parallelogram and $2 \hat{i}+\hat{j}, 4 \hat{i}+5 \hat{j}+4 \hat{k}$ and $-\hat{i}-4 \hat{j}-3 \hat{k}$ be the position vectors of the vertices $A, B$, $D$ respectively. Then the position vector of one of the point of trisection of the diagonal $\mathrm{AC}$ is

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Aicharya · Accepted Answer

Since diagonals of parallelogram bisect each other at mid point.

Hence coordinates of $\mathrm{M}$ $ \begin{aligned} & =\left(\frac{-1+4}{2}, \frac{-4+5}{2}, \frac{-3+4}{2}\right) \\ & =\left(\frac{3}{2}, \frac{1}{2}, \frac{1}{2}\right) \end{aligned} $ Since $M$ is also mid point of point $A$ and $C$ Hence, $\left(\frac{2+x}{2}, \frac{1+y}{2}, \frac{0+z}{2}\right)=\left(\frac{3}{2}, \frac{1}{2}, \frac{1}{2}\right)$ $ \begin{aligned} & \therefore \quad \frac{2+x}{2}=\frac{3}{2} \\ & \Rightarrow x=1 \end{aligned}\left|\Rightarrow \begin{array}{l|l} \frac{1+y}{2}=\frac{1}{2} & \frac{z}{2}=\frac{1}{2} \\ z=0 \end{array}\right| $ Hence coordinate of $c$ is $c(1,0,1)$ Since a trisection divides a line in $1: 2$ and $2: 1$ ratio. When $1: 2$ is in consideration then coordinates of $T_1$ $ \begin{aligned} & =T_1\left(\frac{1(1)+2(2)}{1+2}, \frac{1(0)+2(1)}{1+2}, \frac{1(1)+2(0)}{1+2}\right) \\ & =T_1\left(\frac{5}{3}, \frac{2}{3}, \frac{1}{3}\right) \end{aligned} $ When $2: 1$ is in consideration then co-ordinates of $ \begin{aligned} & T_2=\left(\frac{2(1)+1(2)}{2+1}, \frac{2(0)+1(1)}{2+1}, \frac{2(1)+1(0)}{2+1}\right) \\ & =T_2\left(\frac{4}{3}, \frac{1}{3}, 1\right) \end{aligned} $ Now corresponding to $T_1$, we have position vector $\frac{1}{3}(5 \hat{i}+2 \hat{j}+\hat{k})$. Which is the required vector

Let $A B C D$ be a parallelogram and $2 \hat{i}+\hat{j}, 4 \hat{i}+5 \hat{j}+4 \hat{k}$ and $-\hat{i}-4…

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