Question

If $\overline{r}=(x+y+2)\overline{i}+(2x-y+3)\overline{j}+(x+2y+7)\overline{k}$ where $\overline{r}\cdot \overline{i}=3$, $\overline{r}\cdot \overline{j}=5$ then $\overline{r}\cdot \overline{k}=?$

• A. $4$
• B. $6$
• C. $9$
• D. $8$

Answer 1

The correct option is D $8$

$\begin{array}{c}\overrightarrow{r}\cdot \hat{i}=3\\ \\ x+y+2=3\\ \\ x+y=1----\left(1\right)\\ \\ \overrightarrow{r}\cdot \hat{j}=5\\ \\ 2x-y+3=5\\ \\ 2x-y=2---\left(2\right)\end{array}$

Adding $\left(1\right)$ and $\left(2\right)$

$3x=3$

$x=1$

putting $x=1$ in equation $\left(1\right)$

$x+y=1$

$1+y=1$

$y=0$

Now,

$\overrightarrow{r}\cdot \hat{k}=x+2y+7$

$1+2\times 0+7$

$=8$