Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors such that $|\vec{a}|=1$ such that vectors $3(\vec{a} \times \vec{b} )$ and $2(\vec{b}-(\vec{a} \cdot \vec{b})\vec{a})$ represents $2$ sides of a triangle . If the area of the triangle is $\frac{3}{4} (|\vec{b}|^2+4)$, then find the value of $|\vec{b}|$
Now I used Area of triangle $$=\frac{1}{2} |3[(\vec{a} \times \vec{b} ) ]\times [2(\vec{b}-(\vec{a} \cdot \vec{b})\vec{a})]|$$
which yields $3|-[\vec{b}|^2 \vec{a}+(\vec{a} \cdot \vec{b})^2 \vec{a}]| $ which does not correspond to $\frac{3}{4} (|\vec{b}|^2+4)$. Could someone help me with this?
You found the area is $3|(\vec a\cdot\vec b)^2\vec a - |\vec b|^2\vec a|$. Since $|\vec a|=1$, we have $\vec a\cdot\vec b=|\vec b|\cos\theta$ where $\theta$ is the angle between $\vec a$ and $\vec b$. So area is $$3||\vec b|^2\cos^2\theta-|\vec b|^2|=3|\vec b|^2\sin^2\theta.$$ Forcing it equal to $\frac 34(|\vec b|^2+4)$, we have $$|\vec b|^2=\frac4{4\sin^2\theta-1}.$$ Therefore there are infinite solutions $$|\vec b| = \frac2{\sqrt{4k-1}},\quad\frac14<k\le1,$$ depending on the angle $\theta$ between $\vec a$ and $\vec b$ (where $k=\sin^2\theta$).