Correct Answer - Option 3 : 5
Concept:
Vector triple product
If a, b, and c are three vectors then the vector triple product is written as \(a \times (b \times c)\)
It can be proved that \(a \times (b \times c) = (a.c)b - (a.b)c\)
Calculation:
\(\vec a = \frac {\hat i - 2\hat j}{\sqrt 5}\) ; |a| = 1
\(\overrightarrow a .\overrightarrow a = \frac{{i - 2j}}{{\sqrt 5 }}.\frac{{i - 2j}}{{\sqrt 5 }} = 1\)
\(\vec b = \frac {2\hat i + \hat j + 3\hat k}{\sqrt {14}}\) ; |b| = 1
\(\overrightarrow b .\overrightarrow b = \frac{{2i + j + 3k}}{{\sqrt {14} }}.\frac{{2i + j + 3k}}{{\sqrt {14} }} = 1\)
\(\overrightarrow a .\overrightarrow b = (\frac{{i - 2j}}{{\sqrt 5 }})(\frac{{2i + j + 3k}}{{\sqrt {14} }}) = 0\)
\(\left(2\vec a + \vec b\right)\left[\left(\vec a \times \vec b\right)\times \left(\vec a - 2\vec b\right)\right]\)
\((2\mathop a\limits^ \to + \mathop b\limits^ \to )[(\mathop a\limits^ \to \times \mathop b\limits^ \to ) \times \mathop a\limits^ \to - (\mathop a\limits^ \to \times \mathop b\limits^ \to ) \times \mathop {2b}\limits^ \to ]\)
\((2\mathop a\limits^ \to + \mathop b\limits^ \to )[ - \mathop a\limits^ \to \times (\mathop a\limits^ \to \times \mathop b\limits^ \to ) + \mathop {2b}\limits^ \to \times (\mathop a\limits^ \to \times \mathop b\limits^ \to )]\)
\((2\overrightarrow a + \overrightarrow b )[ - (\overrightarrow a .\overrightarrow b )\overrightarrow a + (\overrightarrow a .\overrightarrow a )\overrightarrow b + (2\overrightarrow b .\overrightarrow b )\overrightarrow a - (2\overrightarrow {b.} \overrightarrow a )\overrightarrow b ]\)
\((2\overrightarrow a + \overrightarrow b )[ - (0)\overrightarrow a + (1)\overrightarrow b + (2)\overrightarrow a - (0)\overrightarrow b ]\)
\((2\overrightarrow a + \overrightarrow b )[\overrightarrow b + 2\overrightarrow a ]\)
\((2\overrightarrow a .\overrightarrow b + \overrightarrow {b.} \overrightarrow b + 4\overrightarrow a .\overrightarrow a + \overrightarrow b .2\overrightarrow a )\)
\(0 + 1 + 4 \times 1 + 0 = 1 + 4 = 5\)
