Question

The vector equation of the line $\frac{5-x}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is _________________.

Answer 1

For given line $\frac{5-x}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is parametric form

$$\begin{gathered} \mathrm{i}.\mathrm{e}.\frac{x-5}{-3}=\frac{y+4}{7}=\frac{z-6}{2} \\ \mathrm{i}.\mathrm{e}.\frac{x-\left(5\right)}{-3}=\frac{y-\left(-4\right)}{7}=\frac{z-\left(6\right)}{2} \end{gathered}$$

Direction ratios are a = −3, b = 7, and c = 2
fixed point is x₁ = 5, x₂ = −4, x₃ = 6
$$\begin{gathered} \therefore \mathrm{Vector}\mathrm{equation}\mathrm{is}\overrightarrow{r}=\left(x_1\hat{i}+x_2\hat{j}+x_3\hat{k}\right)+\lambda \left(a\hat{i}+b\hat{j}+c\hat{k}\right) \\ \mathrm{i}.\mathrm{e}.\overrightarrow{r}=\left(5\hat{i}-4\hat{j}+6\hat{k}\right)+\lambda \left(-3\hat{i}+7\hat{j}+2\hat{k}\right) \end{gathered}$$