Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are not perpendicular to each other and $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{a},\overrightarrow{r}.\overrightarrow{c}=0$, then $\overrightarrow{r}$ is equal to
A. $\overrightarrow{a}-\overrightarrow{c}$
B. $\overrightarrow{b}+\lambda \overrightarrow{a}$, for all scalars $\lambda $
C. $\overrightarrow{b}-\dfrac{\left( \overrightarrow{b}.\overrightarrow{c} \right)}{\left( \overrightarrow{a}-\overrightarrow{c} \right)}\overrightarrow{a}$
D. $\overrightarrow{a}+\overrightarrow{c}$

Answer 1

Hint: In the provided question, we will have to operate the given vector operation, that is, $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{a}$ with arithmetical operations and bring the vectors on the same side of the equation, then we will use the distributive property of vector product over vector addition to relation, $\left( \overrightarrow{r}-\overrightarrow{b} \right)\times \overrightarrow{a}=0$. We can further apply the vector properties to conclude that $\left( \overrightarrow{r}-\overrightarrow{b} \right)$ is parallel to $\overrightarrow{a}$. And thereafter we can operate them mathematically to conclude that, $\overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{a}$.

Complete step-by-step answer:
We are given in the question that, $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{a}$. So, here we will transfer the right hand side or the RHS, $\overrightarrow{b}\times \overrightarrow{a}$ to the left hand side or the LHS, so we get as,
$\overrightarrow{r}\times \overrightarrow{a}-\overrightarrow{b}\times \overrightarrow{a}=0$
We can apply the property of distributive nature of the vector product over addition. So, we can influence the above equation, $\overrightarrow{r}\times \overrightarrow{a}-\overrightarrow{b}\times \overrightarrow{a}=0$ and change it into a more sorted form. Hence, we will get as follows,
$\left( \overrightarrow{r}-\overrightarrow{b} \right)\times \overrightarrow{a}=0$
With our knowledge of vector products, we can conclude that if the cross product of two vectors is resulting into zero, then that implies that those two vectors are parallel to each other. So, this leaves us with the conclusion that, $\left( \overrightarrow{r}-\overrightarrow{b} \right)$ is parallel to $\overrightarrow{a}$. Now, we also know that if two vectors are parallel, then one of the vectors can be represented as some scalar times the second vector. So, using the above statement, we can say that,
 $\left( \overrightarrow{r}-\overrightarrow{b} \right)=\lambda \overrightarrow{a}$, where $\lambda $ is some scalar.
We can arithmetically rearrange the above equation and then we will get the following,
$\overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{a}$

So, the correct answer is “Option B”.

Note: In this question, some students get confused in the initialisation of the problem as they might not know that the cross product of two vectors is equal to zero or they may end up doing some calculation errors.