Mathematics - Vector Question with Solution | TestHub

Given $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos (\frac{\pi }{3})$, we have $1=2|\vec{b}|\frac{1}{2}$, so $|\vec{b}|=1$.

Let $\vec{r}=x\vec{a}+y\vec{b}+z(\vec{a}\times \vec{b})$.

$\vec{r}\cdot \vec{a}=x|\vec{a}{|}^2+y(\vec{a}\cdot \vec{b})=4x+y=2$.

$\vec{r}\cdot \vec{b}=x(\vec{a}\cdot \vec{b})+y|\vec{b}{|}^2=x+y=8$.

Solving these gives $x=-2,y=10$.

Also, $(\vec{r}+2\vec{a}-10\vec{b})\cdot (\vec{a}\times \vec{b})=6$.

Substitute $\vec{r}$: $(-2\vec{a}+10\vec{b}+z(\vec{a}\times \vec{b})+2\vec{a}-10\vec{b})\cdot (\vec{a}\times \vec{b})=6$.

$z(\vec{a}\times \vec{b})\cdot (\vec{a}\times \vec{b})=6$.

$z|\vec{a}\times \vec{b}{|}^2=6$.

$|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin (\frac{\pi }{3})=2\cdot 1\cdot \frac{\sqrt{3}}{2}=\sqrt{3}$.

So, $z(\sqrt{3}{)}^2=6\Rightarrow 3z=6\Rightarrow z=2$.

Thus, $\vec{r}+2\vec{a}-10\vec{b}=(-2\vec{a}+10\vec{b}+2(\vec{a}\times \vec{b}))+2\vec{a}-10\vec{b}=2(\vec{a}\times \vec{b})$.

Comparing with $\vec{r}+2\vec{a}-10\vec{b}=\lambda (\vec{a}\times \vec{b})$, we get $\lambda =2$.