Question

Vector equation of a straight line passing through a point given by position vector $\overline{a}$ parallel to $\overline{b}$ be is given by $\overline{r}=\overline{a}-\lambda \overline{b},where\overline{r}$ , where $\overline{r}$ is r position vector of a general point on the line and $\lambda ϵIR$

• A. True
• B. False

Answer 1

The correct option is A

True

Let’s consider the point $\overline{a}$ and vector $\overline{b}$

The line parallel to $\overline{b}$ and passing through $\overline{a}$ will be something like this

You can clearly see if we shift $\overline{b}$ and do $\overline{a}+\overline{b}$ we get a point on the line.

Moreover, if we keep changing the magnitude of $\overline{b}$ and add it to $\overline{a}$ we will still get a resultant which lies on the line.

Therefore the points of the line required will be $\overline{a}+\lambda \overline{b}$ where $\lambda ϵIR$. Here the value of $\lambda$ is both +ve and –ve. So without loss of generality we can have it as $\overline{a}-\lambda \overline{b}$ where $\lambda ϵIR$. Therefore the given statement is true.