Question

If we have the vectors as $\vec a$ and $\vec b$ in space given by $\vec a = \dfrac{{\hat i - 2\hat j}}{{\sqrt 5 }}$ and $\vec b = \dfrac{{2\hat i + \hat j + 3\hat k}}{{\sqrt {14} }}$, then the value of $\left( {2\vec a + \vec b} \right).\left[ {\left( {\vec a \times \vec b} \right) \times \left( {\vec a - 2\vec b} \right)} \right]$ is
A. 3
B. 4
C. 5
D. 6

Answer 1

In this question, first of all, use the formula of the cross product of three vectors, and then we will proceed by grouping the terms common and simplifying them. Then consider the dot product and mod values of the two vectors $\vec a$ and $\vec b$. Substitute the obtained value in the obtained expression to get the required value.

Complete step-by-step solution:
Given that $\vec a$ and $\vec b$ are vectors in a space given by $\vec a = \dfrac{{\hat i - 2\hat j}}{{\sqrt 5 }}$ and $\vec b = \dfrac{{2\hat i + \hat j + 3\hat k}}{{\sqrt {14} }}$.
Let the given expression is $v = \left( {2\vec a + \vec b} \right).\left[ {\left( {\vec a \times \vec b} \right) \times \left( {\vec a - 2\vec b} \right)} \right]$
By using the formula $\left[ {\left( {\vec x \times \vec y} \right) \times \vec z} \right] = \left[ { - \left( {\vec z.\vec y} \right)\vec x + \left( {\vec z.\vec x} \right)\vec y} \right]$ we have

\Rightarrow v = \left( {2\vec a + \vec b} \right) \cdot \left[ { - \left\\{ {\left( {\vec a - 2\vec b} \right) \cdot \vec b} \right\\}\vec a + \left\\{ {\left( {\vec a - 2\vec b} \right) \cdot \vec a} \right\\}\vec b} \right] \\\ \Rightarrow v = \left( {2\vec a + \vec b} \right) \cdot \left[ { - \left( {\vec a \cdot \vec b - 2\vec b \cdot \vec b} \right)\vec a + \left( {\vec a \cdot \vec a - 2\vec b \cdot \vec a} \right)\vec b} \right] \\\ \Rightarrow v = \left( {2\vec a + \vec b} \right) \cdot \left[ { - \left( {\vec a \cdot \vec b - 2{{\left| {\vec b} \right|}^2}} \right)\vec a + \left( {{{\left| {\vec a} \right|}^2} - 2\vec b \cdot \vec a} \right)\vec b} \right] \\\ \Rightarrow v = \left( {2\vec a + \vec b} \right) \cdot \left[ { - \vec a\left( {\vec a \cdot \vec b} \right) + 2\vec a{{\left| {\vec b} \right|}^2} + {{\left| {\vec a} \right|}^2}\vec b - 2\vec b\left( {\vec a.\vec b} \right)} \right] \\\ \\\

Now consider dot product of the vectors $\vec a$ and $\vec b$

$$\Rightarrow \bar a \cdot \bar b = \left( {\dfrac{{\hat i - 2\hat j}}{{\sqrt 5 }}} \right) \cdot \left( {\dfrac{{2\hat i + \hat j + 3\hat k}}{{\sqrt {14} }}} \right) \\\ \Rightarrow \bar a \cdot \bar b = \left( {\dfrac{{\hat i - 2\hat j}}{{\sqrt 5 }}} \right) \cdot \left( {\dfrac{{2\hat i + \hat j + 3\hat k}}{{\sqrt {14} }}} \right) \\\ \Rightarrow \bar a \cdot \bar b = \dfrac{{2 - 2 + 0}}{{\sqrt {14 \times 5} }} \\\ \therefore \bar a \cdot \bar b = \dfrac{{2 - 2}}{{\sqrt {70} }} = 0 \\\$$

Also, consider the mod values of the vectors $\vec a$ and $\vec b$

$$\Rightarrow \left| {\vec a} \right| = \left| {\dfrac{{\hat i - 2\hat j}}{{\sqrt 5 }}} \right| = \sqrt {\dfrac{{{1^2} + {{\left( { - 2} \right)}^2}}}{{{{\left( {\sqrt 5 } \right)}^2}}}} = \sqrt {\dfrac{{1 + 4}}{5}} = \sqrt {\dfrac{5}{5}} = 1 \\\ \Rightarrow \left| {\vec b} \right| = \left| {\dfrac{{2\hat i + \hat j + 3\hat k}}{{\sqrt {14} }}} \right| = \sqrt {\dfrac{{{2^2} + {1^2} + {3^2}}}{{{{\left( {\sqrt {14} } \right)}^2}}}} = \sqrt {\dfrac{{4 + 1 + 9}}{{14}}} = \sqrt {\dfrac{{14}}{{14}}} = 1 \\\$$

Substituting $\bar a \cdot \bar b = 0,\left| {\bar a} \right| = 1$ and $\left| {\bar b} \right| = 1$, we have

$$\Rightarrow v = \left( {2\vec a + \vec b} \right) \cdot \left[ { - \vec a\left( 0 \right) + 2\vec a{{\left| 1 \right|}^2} + {{\left| 1 \right|}^2}\vec b - 2\vec b\left( 0 \right)} \right] \\\ \Rightarrow v = \left( {2\vec a + \vec b} \right) \cdot \left[ {2\vec a + \vec b} \right] \\\ \Rightarrow v = {\left( {2\vec a + \vec b} \right)^2} = 4{\left| {\vec a} \right|^2} + 4\vec a.\vec b + {\left| {\vec b} \right|^2} \\\$$

Again substituting $\bar a \cdot \bar b = 0,\left| {\bar a} \right| = 1$ and $\left| {\bar b} \right| = 1$, we have

\therefore v = 4 + 0 + 1 = 5 $$ **Thus, the correct option is C. 5** **Note:** Here we have used the algebraic identity $${\left( {\vec a + \vec b} \right)^2} = {\left| {\vec a} \right|^2} + 2\vec a.\vec b + {\left| {\vec b} \right|^2}$$. Always remember that the do product of any two vectors is a scalar and the cross product of two vectors is vector quantity.