Question

Let $\stackrel{\rightharpoonup }{a}=2i-3j+4k,\overrightarrow{b}=7i+j-6k$. If $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{r}\times \overrightarrow{b},\overrightarrow{r}·(i+2j+k)=-3,$ then $\overrightarrow{r}·(2i-3j+k)$ is equal to

• A. $10$
• B. $13$
• C. $12$
• D. $8$

Answer 1

The correct option is C

$12$

Explanation for the correct answer:

Given:

The vectors as $\stackrel{\rightharpoonup }{a}=2i-3j+4k,\overrightarrow{b}=7i+j-6k$.

Then we have, $\stackrel{\rightharpoonup }{a}=(2,-3,4),\overrightarrow{b}=(7,1,-6)$

To find:

The value of $\overrightarrow{r}·(2i-3j+k)$

Explanation:

With the given condition, $\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{r}\times \overrightarrow{b}$, then we get

$\begin{array}{rcl}\overrightarrow{r}\times \overrightarrow{a}-\overrightarrow{r}\times \overrightarrow{b}& =& 0\\ \overrightarrow{r}\times (\overrightarrow{a}-\overrightarrow{b})& =& 0\left[\text{Taking}\overrightarrow{r}\text{common}\right]\\ \left|\overrightarrow{r}\right|\left|\overrightarrow{a}-\overrightarrow{b}\right|\sin \theta & =& 0\end{array}$

Since the cross-product of the vector is zero, we get the angle between the two vectors is zero.

$\begin{array}{rcl}& \Rightarrow & \sin \theta =0\\ \theta & =& 0\end{array}$

Therefore both the vectors lie in the same direction.

Now let us consider, $\overrightarrow{r}=\lambda \left(\overrightarrow{a}-\overrightarrow{b}\right)$ and substitute the values of $\overrightarrow{a}$ and $\overrightarrow{b}$.

$$\begin{gathered} \overrightarrow{r}=\lambda (\overrightarrow{a}-\overrightarrow{b}) \\ \Rightarrow \overrightarrow{r}=\lambda \left(\left(2-7\right)i+\left(-3-1\right)j+\left(4-\left(-6\right)\right)k\right) \\ \Rightarrow \overrightarrow{r}=\lambda \left(-5i+\left(-4\right)j+\left(4+6\right)k\right) \\ \Rightarrow \overrightarrow{r}=\lambda (-5i-4j+10k) \end{gathered}$$

From the given condition, we have $\overrightarrow{r}·(i+2j+k)=-3$.

$\begin{array}{rcl}\lambda \left(-5-8+10\right)& =& -3\\ \lambda (-3)& =& -3\\ \lambda & =& 1\end{array}$

Consider, $\overrightarrow{r}·(2i-3j+k)$ and substitute the required value to simplify.

$\begin{array}{rcl}\overrightarrow{r}·(2i-3j+k)& =& \lambda \left(-5i-4j+10k\right)·(2i-3j+k)\\ & =& \left(1\right)(-5,-4,-10)·(2,-3,1)\\ & =& -10+12+10\\ & =& 12\end{array}$

Therefore, the value of $\overrightarrow{r}·(2i-3j+k)$ is 12.

Hence, the correct option is (C)