$\vec{r}$ and $\vec{s}$ are unit vectors. If $|\vec{r} + \vec{s}| = \sqrt{2}$, find the value of $(4\vec{r} - \vec{s}) \cdot (2\vec{r} + \vec{s})$.

7

The correct answer is Option (3) → 7 ##

Given, $|\vec{r}| = 1, |\vec{s}| = 1$

$(4\vec{r} - \vec{s}) \cdot (2\vec{r} + \vec{s}) = 8|\vec{r}|^2 + 4\vec{r} \cdot \vec{s} - 2\vec{s} \cdot \vec{r} - |\vec{s}|^2$

$= 8(1) + 4|\vec{r}||\vec{s}|\cos \theta - 2|\vec{s}||\vec{r}|\cos \theta - (1)^2$

$=8+4(1)(1)\cos \theta-2(1)(1)\cos \theta-1$

$= 8 + 0 - 0 - 1 \quad [∵\cos \theta = 0 \text{ from above part}]$

$= 7$