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Step 1: Identify the Given Vectors
We have
$ \overrightarrow{a} = 2\widehat{i} + \widehat{j} - 2\widehat{k} $
and
$ \overrightarrow{b} = \widehat{i} + \widehat{j} $.
Step 2: Compute the Magnitude of $ \overrightarrow{a} $
First, let us verify $ \left|\overrightarrow{a}\right| $:
$ \left|\overrightarrow{a}\right|
= \sqrt{(2)^2 + (1)^2 + (-2)^2}
= \sqrt{4 + 1 + 4}
= \sqrt{9}
= 3. $
Step 3: Compute the Cross Product $ \overrightarrow{a} \times \overrightarrow{b} $
Use the determinant form or direct component-wise calculation:
$
\overrightarrow{a} \times \overrightarrow{b}
=
\begin{vmatrix}
\widehat{i} & \widehat{j} & \widehat{k} \\
2 & 1 & -2 \\
1 & 1 & 0
\end{vmatrix}.
$
Carrying out the determinant,
$
\overrightarrow{a} \times \overrightarrow{b}
= (1 \cdot 0 - (-2)\cdot 1)\,\widehat{i}
- (2\cdot 0 - (-2)\cdot 1)\,\widehat{j}
+ (2\cdot 1 - 1\cdot 1)\,\widehat{k}.
$
Simplify each component:
$
= (0 + 2)\,\widehat{i}
- (0 + 2)\,\widehat{j}
+ (2 - 1)\,\widehat{k}
= 2\widehat{i} - 2\widehat{j} + \widehat{k}.
$
Step 4: Compute the Magnitude of $ \overrightarrow{a} \times \overrightarrow{b} $
$
\left| \overrightarrow{a} \times \overrightarrow{b} \right|
= \sqrt{(2)^2 + (-2)^2 + (1)^2}
= \sqrt{4 + 4 + 1}
= \sqrt{9}
= 3.
$
Step 5: Use the Condition $ \left| (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \right| = 3 $ with Angle $30^\circ$
We know the magnitude of a cross product of two vectors $ \overrightarrow{u} $ and $ \overrightarrow{v} $ is
$ \left|\overrightarrow{u}\right| \,\left|\overrightarrow{v}\right| \sin \theta $.
Here, $ \overrightarrow{u} = \overrightarrow{a} \times \overrightarrow{b} $ and $ \overrightarrow{v} = \overrightarrow{c} $, and the angle between $ \overrightarrow{c} $ and $ \overrightarrow{a} \times \overrightarrow{b} $ is $ 30^\circ $. Thus,
$
\bigl| (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \bigr|
= \left|\overrightarrow{a} \times \overrightarrow{b}\right| \,\left|\overrightarrow{c}\right| \sin 30^\circ.
$
Given $ \bigl| (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \bigr| = 3 $ and $ \left|\overrightarrow{a} \times \overrightarrow{b} \right| = 3 $, we get
$
3 = 3 \,\left|\overrightarrow{c}\right| \times \sin 30^\circ.
$
But $ \sin 30^\circ = \frac{1}{2} $, so
$
3 = 3 \,\left|\overrightarrow{c}\right| \times \frac{1}{2}
\quad\Longrightarrow\quad
3 = \frac{3}{2}\,\left|\overrightarrow{c}\right|
\quad\Longrightarrow\quad
\left|\overrightarrow{c}\right| = 2.
$
Step 6: Use the Condition $ \left|\overrightarrow{c} - \overrightarrow{a}\right| = 3 $
We are given
$ \left|\overrightarrow{c} - \overrightarrow{a}\right| = 3. $
Squaring both sides:
$
\left|\overrightarrow{c} - \overrightarrow{a}\right|^2 = 3^2 = 9.
$
Now,
$
\left|\overrightarrow{c} - \overrightarrow{a}\right|^2
= \left|\overrightarrow{c}\right|^2
+ \left|\overrightarrow{a}\right|^2
- 2\,\overrightarrow{c}\cdot\overrightarrow{a}.
$
Substitute $ \left|\overrightarrow{c}\right| = 2 $ and $ \left|\overrightarrow{a}\right| = 3 $:
$
(2)^2 + (3)^2 - 2 (\overrightarrow{c}\cdot\overrightarrow{a}) = 9.
$
So,
$
4 + 9 - 2(\overrightarrow{c}\cdot\overrightarrow{a}) = 9
\quad\Longrightarrow\quad
13 - 2(\overrightarrow{c}\cdot\overrightarrow{a}) = 9
\quad\Longrightarrow\quad
2(\overrightarrow{c}\cdot\overrightarrow{a}) = 4
\quad\Longrightarrow\quad
\overrightarrow{c}\cdot\overrightarrow{a} = 2.
$
Step 7: Conclusion
Hence, $ \overrightarrow{a} \cdot \overrightarrow{c} = 2 $.
