Question

ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, $4\hat{i}+5\hat{j}-10\hat{k},2\hat{i}-3\hat{j}+4\hat{k}\mathrm{and}-\hat{i}+2\hat{j}+\hat{k}.$ Find the vector equation of the line BD. Also, reduce it to cartesian form.

Answer 1

We know that the position vector of the mid-point of $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{\overrightarrow{a}+\overrightarrow{b}}{2}$.

Let the position vector of point D be $x\hat{i}+y\hat{j}+z\hat{k}$.

Position vector of mid-point of A and C = Position vector of mid-point of B and D

$$\begin{gathered} \therefore \frac{\left(4\hat{i}+5\hat{j}-10\hat{k}\right)+\left(-\hat{i}+2\hat{j}+\hat{k}\right)}{2}=\frac{\left(2\hat{i}-3\hat{j}+4\hat{k}\right)+\left(x\hat{i}+y\hat{j}+z\hat{k}\right)}{2} \\ \Rightarrow \frac{3}{2}\hat{i}+\frac{7}{2}\hat{j}-\frac{9}{2}\hat{k}=\left(\frac{x+2}{2}\right)\hat{i}+\left(\frac{-3+y}{2}\right)\hat{j}+\left(\frac{4+z}{2}\right)\hat{k} \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficient}\mathrm{of}\hat{i},\hat{j}\mathrm{and}\hat{k},\mathrm{we}\mathrm{get} \\ \frac{x+2}{2}=\frac{3}{2} \\ \Rightarrow x=1 \\ \frac{-3+y}{2}=\frac{7}{2} \\ \Rightarrow y=10 \\ \frac{4+z}{2}=-\frac{9}{2} \\ \Rightarrow z=-13 \\ \mathrm{Position}\mathrm{vector}\mathrm{of}\mathrm{point}D=\hat{i}+10\hat{j}-13\hat{k} \end{gathered}$$

The vector equation of line BD passing through the points with position vectors $\overrightarrow{a}$(B) and $\overrightarrow{b}$(D) is $\overrightarrow{r}=\overrightarrow{a}+\lambda \left(\overrightarrow{b}-\overrightarrow{a}\right)$.

Here,
$$\begin{gathered} \overrightarrow{a}=2\hat{i}-3\hat{j}+4\hat{k} \\ \overrightarrow{b}=\hat{i}+10\hat{j}-13\hat{k} \end{gathered}$$

Vector equation of the required line is
$$\begin{gathered} \overrightarrow{r}=\left(2\hat{i}-3\hat{j}+4\hat{k}\right)+\lambda \left\{\left(\hat{i}+10\hat{j}-13\hat{k}\right)-\left(2\hat{i}-3\hat{j}+4\hat{k}\right)\right\} \\ \Rightarrow \overrightarrow{r}=\left(2\hat{i}-3\hat{j}+4\hat{k}\right)+\lambda \left(-\hat{i}+13\hat{j}-17\hat{k}\right)...\left(1\right) \\ \mathrm{Here},\lambda \mathrm{is}\mathrm{a}\mathrm{parameter}. \end{gathered}$$

Reducing (1) to cartesian form, we get

$$\begin{gathered} x\hat{i}+y\hat{j}+z\hat{k}=\left(2\hat{i}-3\hat{j}+4\hat{k}\right)+\lambda \left(-\hat{i}+13\hat{j}-17\hat{k}\right)[\mathrm{Putting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\mathrm{in}(1\left)\right] \\ \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left(2-\lambda \right)\hat{i}+\left(-3+13\lambda \right)\hat{j}+\left(4-17\lambda \right)\hat{k} \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}\hat{i},\hat{j}\mathrm{and}\hat{k},\mathrm{we}\mathrm{get} \\ x=2-\lambda ,y=-3+13\lambda ,z=4-17\lambda \\ \Rightarrow \frac{x-2}{-1}=\lambda ,\frac{y+3}{13}=\lambda ,\frac{z-4}{-17}=\lambda \\ \Rightarrow \frac{x-2}{-1}=\frac{y+3}{13}=\frac{z-4}{-17}=\lambda \\ \Rightarrow \frac{x-2}{1}=\frac{y+3}{-13}=\frac{z-4}{17}=-\lambda \\ \mathrm{Hence},\mathrm{the}\mathrm{cartesian}\mathrm{form}\mathrm{of}\left(1\right)\mathrm{is} \\ \frac{x-2}{1}=\frac{y+3}{-13}=\frac{z-4}{17} \end{gathered}$$