<s>[INST] 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.
 [/INST] 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.
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