If \(\vec{r}\) and \(\vec{s}\) are unit vectors such that \(|\vec{r}+\vec{s}| = \sqrt{3}\), then the value of \((3\vec{r}-2\vec{s})\cdot(\vec{r}+4\vec{s})\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Since \(|\vec{r}| = |\vec{s}| = 1\), squaring the given magnitude gives \(\vec{r}\cdot\vec{s}\):
\(|\vec{r}+\vec{s}|^2 = |\vec{r}|^2+|\vec{s}|^2+2\,\vec{r}\cdot\vec{s}\)
\(3 = 1+1+2\,\vec{r}\cdot\vec{s}\)
\( \displaystyle \vec{r}\cdot\vec{s} = \frac{1}{2} \)
Expanding by the distributive law of the scalar product:
\((3\vec{r}-2\vec{s})\cdot(\vec{r}+4\vec{s}) = 3|\vec{r}|^2+12\,\vec{r}\cdot\vec{s}-2\,\vec{s}\cdot\vec{r}-8|\vec{s}|^2\)
\( = 3+10\,\vec{r}\cdot\vec{s}-8\)
\( \displaystyle = 3+10\times\frac{1}{2}-8 = 0 \)