Question

Let $a=i+j$ and $b=2i-k$, then the point of intersection of the lines $r\times a=b\times a$ and $r\times b=a\times b$ is
A. $-i+j+k$
B. $3i-j+k$
C. $3i+j-k$
D. $i-j-k$

Answer 1

Hint: In the question, we have been given two vectors and two lines, which can be simplified using different properties. Since the first line here is $r\times a=b\times a$, it can be written as, $\overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{a}$ and the other line, $r\times b=a\times b$ can be written as, $\overrightarrow{r}=\overrightarrow{a}+\mu \overrightarrow{b}$, where $\lambda $ and $\mu $ are the constants and for intersection, $\overrightarrow{r}=\overrightarrow{r}$, we obtain $\lambda =\mu =1$. We will thus proceed further by solving it and finding the value of $\overrightarrow{r}$.

Complete step by step answer:
In this question, we have been given the values of two vectors as, $a=i+j$ and $b=2i-k$ and we have been asked to find the point of intersection of the lines, $r\times a=b\times a$ and $r\times b=a\times b$. Before we proceed, we need to understand the concept and the different types of representation of a line in vectors. So, if we take the line in the vector form, that is, $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{a}$, then $\overrightarrow{a}$ is the position vector, and so it can be written as, $\overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{a}$, where $\lambda $ is a constant. Similarly if we take the other line, $\overrightarrow{r}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{b}$, then $\overrightarrow{b}$ is the position vector, and it can be written as $\overrightarrow{r}=\overrightarrow{a}+\mu \overrightarrow{b}$, where $\mu $ is the constant. Now, for the intersection of both the lines, $\overrightarrow{r}=\overrightarrow{r}$. So, we will get,
$\overrightarrow{b}+\lambda \overrightarrow{a}=\overrightarrow{a}+\mu \overrightarrow{b}$
So, on comparing the coefficients, we will get,
$\lambda =\mu =1$
As the coefficients are the same, we get the point of intersection as,
$\overrightarrow{r}=\overrightarrow{a}+\overrightarrow{b}$
Now we have been given the values of $a=i+j$ and $b=2i-k$, so we will substitute them and get the value of $\overrightarrow{r}$,
$\begin{align}
  & \overrightarrow{r}=i+j+2i-k \\
 & \overrightarrow{r}=3i+j-k \\
\end{align}$
Therefore, we get the correct option as option C.

Note:
 We have to remember that and keep in mind that $\overrightarrow{r}$ acts as the point of intersection when a point vector is asked. To understand the conversion better, let us consider $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{a}$. Now, we will rearrange it as
$\begin{align}
  & \overrightarrow{r}\times \overrightarrow{a}-\overrightarrow{b}\times \overrightarrow{a}=0 \\
 & \overrightarrow{a}\times \left( \overrightarrow{r}-\overrightarrow{b} \right)=0 \\
\end{align}$
Now, it is in the form of cross product of two vectors = 0. Now we know that this happens when two vectors are parallel, so we have vector $\overrightarrow{a}$ parallel to $\left( \overrightarrow{r}-\overrightarrow{b} \right)$ . Parallel vectors are related by a constant, so we can write that
$\begin{align}
  & \overrightarrow{r}-\overrightarrow{b}=\lambda \overrightarrow{a} \\
 & \overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{a} \\
\end{align}$
Similarly, we can do it for $r\times b=a\times b$ too.