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 $|2\vec a+3\vec b+4\vec c|$

Question

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.

Answer 1

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.

Answer 2

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 $\sqrt{176 - 16 - 96} = 8$.