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 `vec a` and parallel to vector \[\overrightarrow{b}\] is \[\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}\]   . Here,

\[\overrightarrow{a} = 5 \hat {i} + 2 \hat{j} - 4 \hat{k}\]

\[ \overrightarrow{b} = 3 \hat{i} + 2 \hat{j} - 8 \hat{k}\]

Vector equation of the required line is given by

\[\overrightarrow{r} = \left( 5 \hat{i} + 2 \hat{j}- 4 \hat{k} \right) + \lambda \left( 3 \hat{i} + 2 \hat{j} - 8 \hat{k} \right) . . . \left( 1 \right) \]

\[ \text{ Here }, \lambda \text{ is a parameter } . \]

Reducing (1) to cartesian form, we get

\[x \hat{i} + y \hat{j} + z \hat{k} = \left( 5 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda \left( 3 \hat{i} + 2 \hat{j} - 8 \hat{k} \right) [\text{ Putting }  \ \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}  \  \in (1)]\]

\[ \Rightarrow x \hat{i} + y \hat{j} + z \hat{k} = \left( 5 + 3\lambda \right) \hat{i} + \left( 2 + 2 \lambda \right) \hat{j} + \left( - 4 - 8 \lambda \right) \hat{k} \]

\[\text{ Comparing the coefficients of }  \hat{i} , \hat{j}  \text{and } \hat{k} , \text{ we get }\]

\[x = 5 + 3\lambda, y = 2 + 2 \lambda, z = - 4 - 8 \lambda\]

\[ \Rightarrow \frac{x - 5}{3} = \lambda, \frac{y - 2}{2} = \lambda, \frac{z + 4}{- 8} = \lambda\]

\[ \Rightarrow \frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{- 8} = \lambda\]

\[\text{ Hence, the cartesian form of (1) is } \]

\[\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{- 8}\]