Question

Let, $\overrightarrow{a}=\stackrel{⏜}{i}+\alpha \stackrel{⏜}{j}+3\stackrel{⏜}{k},\overrightarrow{b}=\stackrel{⏜}{3i}-\alpha \stackrel{⏜}{j}+\stackrel{⏜}{k}$. The area of the parallelogram whose adjacent sides are represented as $\overrightarrow{a},\overrightarrow{b}$ is $8\sqrt{3}$ square units, then $\overrightarrow{a}.\overrightarrow{b}$is equal to

Answer 1

Find $\overrightarrow{a}.\overrightarrow{b}$and $\overrightarrow{a}\times \overrightarrow{b}$:

As we know,

The area of the parallelogram $\begin{array}{rcl}& =& \left|\overrightarrow{a}\times \overrightarrow{b}\right|\end{array}$

$=8\sqrt{3}$

$\Rightarrow$ $\begin{array}{rcl}\left|(\stackrel{⏜}{i}+\alpha \stackrel{⏜}{j}+3\stackrel{⏜}{k)}\times (\stackrel{⏜}{3i}-\alpha \stackrel{⏜}{j}+\stackrel{⏜}{k})\right|& =& 8\sqrt{3}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\left|(-3\alpha k+9j-\alpha k+3\alpha i-j+\alpha i)\right|& =& 8\sqrt{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\left|\left(4\alpha \right)i+8j-\left(4\alpha \right)k\right|& =& 8\sqrt{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\sqrt{(16{\alpha }^2+64+16{\alpha }^2)}& =& 8\sqrt{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}32{\alpha }^2+64& =& 192\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\alpha }^2& =& 4\end{array}$

Now,

$\begin{array}{rcl}\overrightarrow{a}.\overrightarrow{b}& =& (3-{\alpha }^2+3)\\ & =& (6-4)\\ & =& 2\end{array}$

$\therefore$the area of parallelogram is $2$ square units