If $\vec a,\vec b,\vec c$ be three vectors such that $|\vec a|=1,|\vec b|=2,|\vec c|=4$ and then find the value of $|2\vec a+3\vec b+4\vec c|$ If $\vec a,\vec b,\vec c$ be three vectors such that
$\vert \vec a\vert =1,\vert \vec b\vert =2,\vert \vec c\vert=4$
and
$\vec a \cdot \vec b+\vec b \cdot \vec c+\vec c \cdot\vec a=-10$
then find the value of $\vert 2\vec a+3\vec b+4\vec c \vert$
My Attempt
$\vert \vec a+\vec b+\vec c \vert^2=a^2+b^2+c^2+2(\vec a \cdot \vec b+\vec b \cdot \vec c+\vec c \cdot \vec a)=1+4+16-20=1$
So, $\vert \vec a+\vec b+\vec c \vert=1$
Further by hit and trial I could see that if $\vec a=\vec i,\vec b=2\vec i,\vec c=-4\vec i$ (where $\vec i$ is unit vector along x-axis) satisfies all conditions.
So, $\vert 2\vec a+3\vec b+4\vec c \vert =\vert 2\vec i+6\vec i-16\vec i\vert =8$
But can there be a better way to do this.
 A: Recall that $\|\vec x\|^2 = \vec x \cdot \vec x$. Then
$$\begin{align*}
\|2\vec a + 3\vec b + 4\vec c\|^2 &= (2\vec a + 3\vec b + 4\vec c) \cdot (2\vec a + 3\vec b + 4\vec c) \\
&= 4\|\vec a\|^2 + 9\|\vec b\|^2 + 16\|\vec c\|^2 + 2 \left(6\vec a\cdot\vec b + 8\vec a\cdot\vec c + 12\vec b\cdot \vec c\right)\\
&= (4\cdot1)+(9\cdot4)+16^2 + (12\cdot(-10)) + 4\vec a\cdot\vec c + 12\vec b\cdot\vec c \\
&= 176 + 4 (\vec a\cdot \vec c + 3\vec b \cdot \vec c) \\
&= 176 + 16 \cos(\theta) + 96 \cos(\phi)
\end{align*}$$
where $\theta$ is the angle between $\vec a$ and $\vec c$, and $\phi$ is the angle between $\vec b$ and $\vec c$. If we take $\theta=\phi=\pi$ as you've done, the minimum value of $\|2\vec a+3\vec b + 4\vec c\|$ is $\sqrt{176 - 16 - 96} = 8$.
A: You have
$$2a + 3b +4c = 3(a+b+c) -a+c.$$ therefore
$$\begin{aligned}
\lVert 2a + 3b +4c \rVert^2 &= 9 \lVert a + b +c \rVert^2 + \lVert a \rVert^2 + \lVert c \rVert^2 - 6 \lVert a \rVert^2 - 6 a \cdot b - 6 a \cdot c + 6 \lVert c \rVert^2 + 6 b \cdot c + 6 a \cdot c\\
&=9 + 1 + 16 -6 + 96  - 6 a \cdot b + 6 a \cdot c\\
&=116   - 6 a \cdot b + 6 b \cdot c
\end{aligned}$$
... and the problem is not fully determined.
