Question

Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector $3\hat{i}+2\hat{j}-8\hat{k}.$

Answer 1

We know that the vector equation of a line passing through a point with position vector $\overrightarrow{a}$ and parallel to vector $\overrightarrow{b}$ is $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$.

Here,
$$\begin{gathered} \overrightarrow{a}=5\hat{i}+2\hat{j}-4\hat{k} \\ \overrightarrow{b}=3\hat{i}+2\hat{j}-8\hat{k} \end{gathered}$$

Vector equation of the required line is given by
$$\begin{gathered} \overrightarrow{r}=\left(5\hat{i}+2\hat{j}-4\hat{k}\right)+\lambda \left(3\hat{i}+2\hat{j}-8\hat{k}\right)...\left(1\right) \\ \mathrm{Here},\lambda \mathrm{is}\mathrm{a}\mathrm{parameter}. \end{gathered}$$

Reducing (1) to cartesian form, we get

$$\begin{gathered} x\hat{i}+y\hat{j}+z\hat{k}=\left(5\hat{i}+2\hat{j}-4\hat{k}\right)+\lambda \left(3\hat{i}+2\hat{j}-8\hat{k}\right)[\mathrm{Putting}\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}\mathrm{in}(1\left)\right] \\ \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=\left(5+3\lambda \right)\hat{i}+\left(2+2\lambda \right)\hat{j}+\left(-4-8\lambda \right)\hat{k} \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}\mathrm{of}\hat{i},\hat{j}\mathrm{and}\hat{k},\mathrm{we}\mathrm{get} \\ x=5+3\lambda ,y=2+2\lambda ,z=-4-8\lambda \\ \Rightarrow \frac{x-5}{3}=\lambda ,\frac{y-2}{2}=\lambda ,\frac{z+4}{-8}=\lambda \\ \Rightarrow \frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}=\lambda \\ \mathrm{Hence},\mathrm{the}\mathrm{cartesian}\mathrm{form}\mathrm{of}\left(1\right)\mathrm{is} \\ \frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8} \end{gathered}$$