Question

If $\overrightarrow{a}=\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{2j}}\,+\overset{\wedge }{\mathop{3k}}\,$ and $\overset{\to }{\mathop{b}}\,=\overset{\wedge }{\mathop{i}}\,\times \left( a\times \overset{\wedge }{\mathop{i}}\, \right)+\overset{\wedge }{\mathop{j}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{j}}\, \right)+\overset{\wedge }{\mathop{k}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{k}}\, \right)$ then length of b is equal to:
A. $\sqrt{12}$
B.$2\sqrt{12}$
C.$3\sqrt{14}$
D.$2\sqrt{14}$

Answer 1

Hint: Use vector triple product concept.

Here, we have vectors given
$\overrightarrow{a}=\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{2j}}\,+\overset{\wedge }{\mathop{3k}}\,............\left( 1 \right)$
And
$\overset{\to }{\mathop{b}}\,=\overset{\wedge }{\mathop{i}}\,\times \left( a\times \overset{\wedge }{\mathop{i}}\, \right)+\overset{\wedge }{\mathop{j}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{j}}\, \right)+\overset{\wedge }{\mathop{k}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{k}}\, \right)...........\left( 2 \right)$
We need to find the length of the vector $\overset{\to }{\mathop{b}}\,$. Vector $\overset{\to }{\mathop{b}}\,$is not in the generalized form $x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{k}}\,,$ so first we need to convert $\overset{\to }{\mathop{b}}\,$to general form of vector as mentioned above.
We know length of any vector$\overset{\to }{\mathop{A}}\,=x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,+z\overset{\wedge }{\mathop{k}}\,$ is magnitude of this vector which is given by
$\left| \overset{\to }{\mathop{A}}\, \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}..............\left( 3 \right)$
Now, we can observe from $\overset{\to }{\mathop{b}}\,$ that it is summation of three triple product vectors i.e.
$\overset{\wedge }{\mathop{i}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{i}}\, \right),\overset{\wedge }{\mathop{j}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{j}}\, \right),\overset{\wedge }{\mathop{k}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{k}}\, \right)$
We have vector triple product formula as
$\overset{\to }{\mathop{A}}\,\times \left( \overset{\to }{\mathop{B}}\,\times \overset{\to }{\mathop{C}}\, \right)=\left( \overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{C}}\, \right)\overset{\to }{\mathop{B}}\,-\left( \overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\, \right)\overset{\to }{\mathop{C}}\,..........\left( 4 \right)$
We can prove the above formula by taking three generalized vectors and solving LHS and RHS both.
Let us apply the vector triple product formula to $\overset{\to }{\mathop{b}}\,$.
We have
$\overset{\to }{\mathop{b}}\,=\overset{\wedge }{\mathop{i}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{i}}\, \right)+\overset{\wedge }{\mathop{j}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{j}}\, \right)+\overset{\wedge }{\mathop{k}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{k}}\, \right)$
Let us suppose
\[\begin{align}
  & \overset{\to }{\mathop{m}}\,=\overset{\wedge }{\mathop{i}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{i}}\, \right) \\
 & \overset{\to }{\mathop{n}}\,=\overset{\wedge }{\mathop{j}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{j}}\, \right) \\
 & \overset{\to }{\mathop{p}}\,=\overset{\wedge }{\mathop{k}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{k}}\, \right) \\
\end{align}\]
Therefore we can write $\overset{\to }{\mathop{b}}\,$in for of $\overset{\to }{\mathop{m}}\,,\overset{\to }{\mathop{n}}\,\And \overset{\to }{\mathop{p}}\,$ as $\overset{\to }{\mathop{b}}\,=\overset{\to }{\mathop{m}}\,+\overset{\to }{\mathop{n}}\,+\overset{\to }{\mathop{p}}\,.............\left( 5 \right)$
Now, Let us apply vector triple product formula from equation (4) to simplify $\overset{\to }{\mathop{m}}\,,\overset{\to }{\mathop{n}}\,\And \overset{\to }{\mathop{p}}\,$.
We need to know dot product of two vectors as well to solve above expression which is given by
$\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\,=\left| \overset{\to }{\mathop{A}}\, \right|\left| \overset{\to }{\mathop{B}}\, \right|\cos \theta ....................\left( 6 \right)$
$\theta $ is the angle between $\overset{\to }{\mathop{A}}\,\And \overset{\to }{\mathop{B}}\,$ .
The angle between i and i (or the same vector) is 0. And angles between (i, j), (j, k) or (i, k) is $90{}^\circ $ which gives product 0.

Where, i, j, k are unit vector i.e. $\left| \overset{\wedge }{\mathop{i}}\, \right|=\left| \overset{\wedge }{\mathop{j}}\, \right|=\left| \overset{\wedge }{\mathop{k}}\, \right|=1$
$\begin{align}
  & \overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{i}}\,=\left| \overset{\wedge }{\mathop{i}}\, \right|\left| \overset{\wedge }{\mathop{i}}\, \right|\cos 0{}^\circ =1,\text{ similarly, }\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{j}}\,=1\And \overset{\wedge }{\mathop{k}}\,.\overset{\wedge }{\mathop{k}}\,=1 \\
 & \overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{j}}\,=\left| \overset{\wedge }{\mathop{i}}\, \right|\left| \overset{\wedge }{\mathop{j}}\, \right|\cos 90{}^\circ =0,\text{similarly, }\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{k}}\,=0\And \overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{k}}\,=0 \\
\end{align}$
Therefore, we can write $\overset{\to }{\mathop{m}}\,,\overset{\to }{\mathop{n}}\,,\overset{\to }{\mathop{p}}\,$ as
$\overset{\to }{\mathop{m}}\,=\overset{\wedge }{\mathop{i}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{i}}\, \right)=\left( \overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{i}}\, \right)\overset{\to }{\mathop{a}}\,-\left( \overset{\wedge }{\mathop{i}}\,.\overset{\to }{\mathop{a}}\, \right)\overset{\wedge }{\mathop{i}}\,$
Substitute $\overset{\to }{\mathop{a}}\,=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,$ from equation (1),
We get
\[\overset{\to }{\mathop{m}}\,=\left( \overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{i}}\, \right)\left( \overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right)-\overset{\wedge }{\mathop{i}}\,.\left( \overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right)\overset{\wedge }{\mathop{i}}\,\]
$\begin{align}
  & m=\left| i \right|\left| i \right|\cos 0{}^\circ \left( i+2j+3\overset{\wedge }{\mathop{k}}\, \right)-\left( \overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{k}}\, \right) \\
 & m=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,-\left( 1+0+0 \right)i \\
 & m=2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,........................\left( 7 \right) \\
\end{align}$
Now coming to $\overset{\to }{\mathop{n}}\,,$ we get
$\overset{\to }{\mathop{n}}\,=\overset{\wedge }{\mathop{j}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{j}}\, \right)=\left( \overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{j}}\, \right)\overset{\to }{\mathop{a}}\,-\left( \overset{\wedge }{\mathop{j}}\,-\overset{\to }{\mathop{a}}\, \right)\left( \overset{\wedge }{\mathop{j}}\, \right)$
Substitute $\overset{\to }{\mathop{a}}\,=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,$ from equation (1),
$\overset{\to }{\mathop{n}}\,=\left( \overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{j}}\, \right)\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,-\overset{\wedge }{\mathop{j}}\,.\left( \overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right)\overset{\wedge }{\mathop{j}}\,$
We can simplify $\overset{\to }{\mathop{n}}\,,$as
$\begin{align}
  & \overset{\to }{\mathop{n}}\,=1\left( \overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{i}}\,+3\overset{\wedge }{\mathop{k}}\, \right)-\left( 0+2\left( 1 \right)+3\left( 0 \right) \right)\overset{\wedge }{\mathop{j}}\, \\
 & \overset{\to }{\mathop{n}}\,=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,-2\overset{\wedge }{\mathop{j}}\, \\
 & \overset{\to }{\mathop{n}}\,=\overset{\wedge }{\mathop{i}}\,+3\overset{\wedge }{\mathop{k}}\,.....................\left( 8 \right) \\
\end{align}$
Similarly we can write $\overset{\to }{\mathop{p}}\,$ as;
$\overset{\to }{\mathop{p}}\,=\overset{\wedge }{\mathop{k}}\,\times \left( \overset{\to }{\mathop{a}}\,\times \overset{\wedge }{\mathop{k}}\, \right)=\overset{\wedge }{\mathop{k}}\,.\overset{\wedge }{\mathop{k}}\,\overset{\to }{\mathop{a}}\,-\left( \overset{\wedge }{\mathop{k}}\,-\overset{\to }{\mathop{a}}\, \right)\overset{\wedge }{\mathop{k}}\,$
Substituting $\overrightarrow{a}=\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{2j}}\,+\overset{\wedge }{\mathop{3k}}\,$ from equation (1) we can rewrite $\overset{\to }{\mathop{p}}\,$;
$\begin{align}
  & \overset{\to }{\mathop{p}}\,=\left( \overset{\wedge }{\mathop{k}}\,.\overset{\wedge }{\mathop{k}}\, \right)\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,-\overset{\wedge }{\mathop{k}}\,.\left( \overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right)\overset{\wedge }{\mathop{k}}\, \\
 & \overset{\to }{\mathop{p}}\,=1\left( \overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\, \right)-\left( 0+2\left( 0 \right)+3\left( 1 \right) \right)\overset{\wedge }{\mathop{k}}\, \\
 & \overset{\to }{\mathop{p}}\,=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,-3\overset{\wedge }{\mathop{k}}\, \\
 & \overset{\to }{\mathop{p}}\,=\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,.............\left( 9 \right) \\
\end{align}$
We have,
$\overset{\to }{\mathop{b}}\,=\overset{\to }{\mathop{m}}\,+\overset{\to }{\mathop{n}}\,+\overset{\to }{\mathop{p}}\,$ from equation (5) and now substitute $\overset{\to }{\mathop{m}}\,,\overset{\to }{\mathop{n}}\,,\overset{\to }{\mathop{p}}\,$ from equations (7),(8),(9), we get;
\[\begin{align}
  & \overset{\to }{\mathop{b}}\,=2\overset{\wedge }{\mathop{j}}\,+3\overset{\wedge }{\mathop{k}}\,+\overset{\wedge }{\mathop{i}}\,+3\overset{\wedge }{\mathop{k}}\,+\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\, \\
 & \overset{\to }{\mathop{b}}\,=2\overset{\wedge }{\mathop{j}}\,+4\overset{\wedge }{\mathop{j}}\,+6\overset{\wedge }{\mathop{k}}\, \\
\end{align}\]
Now, coming to the question we need to find the length of $\overset{\to }{\mathop{b}}\,$or magnitude. Since, magnitude of any vector is given in equation (3),
Therefore, length of $\overset{\to }{\mathop{b}}\,$is
$\begin{align}
  & \left| \overset{\to }{\mathop{b}}\, \right|=\sqrt{{{2}^{2}}+{{4}^{2}}+{{6}^{2}}} \\
 & \left| \overset{\to }{\mathop{b}}\, \right|=\sqrt{4+16+36} \\
 & \left| \overset{\to }{\mathop{b}}\, \right|=\sqrt{56}=\sqrt{4\times 14} \\
 & \left| \overset{\to }{\mathop{b}}\, \right|=2\sqrt{14} \\
\end{align}$
Length of $\overset{\to }{\mathop{b}}\,$is $\sqrt{56}\text{ or }2\sqrt{14}$ .

Hence option D is the correct answer.

Note: Another approach for this question would be that we can put $\overset{\to }{\mathop{a}}\,$in $\overset{\to }{\mathop{b}}\,$and then find out cross products such as
$\overset{\to }{\mathop{b}}\,=i\times \left( \overset{\to }{\mathop{a}}\,\times i \right)+j\times \left( a\times j \right)+\overset{\wedge }{\mathop{k}}\,\times \left( a\times \overset{\wedge }{\mathop{k}}\, \right)$
Substitute$\text{ }\overset{\to }{\mathop{a}}\,=i+2j+3\overset{\wedge }{\mathop{k}}\,$
$\overset{\to }{\mathop{b}}\,=i\times \left( \left( i\times 2j\times 3\overset{\wedge }{\mathop{k}}\, \right)\times i \right)+j\times \left( \left( i+2j\times 3\overset{\wedge }{\mathop{k}}\, \right)\times \overset{\wedge }{\mathop{j}}\, \right)+\overset{\wedge }{\mathop{k}}\,\times \left( \left( i+2j\times 3\overset{\wedge }{\mathop{k}}\, \right)\times \overset{\wedge }{\mathop{k}}\, \right)$
Now, we can use formula of cross product of
$\overset{\to }{\mathop{A}}\,={{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}\overset{\wedge }{\mathop{k}}\,\And \overset{\to }{\mathop{B}}\,={{b}_{1}}i+{{b}_{2}}j+{{b}_{3}}\overset{\wedge }{\mathop{k}}\,$ as
$\begin{align}
  & \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\,={{\left| \begin{matrix}
   i & j & \overset{\wedge }{\mathop{k}}\, \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|}^{-}} \\
 & =\left( {{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}} \right)\overset{\wedge }{\mathop{i}}\,-\left( {{a}_{1}}{{b}_{3}}-{{b}_{1}}{{a}_{3}} \right)\overset{\wedge }{\mathop{j}}\,+\left( {{a}_{1}}{{b}_{2}}-{{b}_{1}}{{a}_{2}} \right)\overset{\wedge }{\mathop{k}}\, \\
\end{align}$
Use the above relation to simplify $\overset{\to }{\mathop{b}}\,$.
Above method would be longer than the given solution,
One can go wrong while using formula $\overset{\to }{\mathop{A}}\,\times \left( \overset{\to }{\mathop{B}}\,\times \overset{\to }{\mathop{C}}\, \right)=\left( \overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{C}}\, \right)\overset{\to }{\mathop{B}}\,-\left( \overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\, \right)\overset{\to }{\mathop{C}}\,$
Where $\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{C}}\,\And \overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\,$ are constants and multiplied with $\overset{\to }{\mathop{B}}\,\And \overset{\to }{\mathop{C}}\,$ respectively.
And
$\begin{align}
  & \left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\, \right)\times \overset{\to }{\mathop{C}}\,=-\overset{\to }{\mathop{C}}\,\times \left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\, \right) \\
 & =-\left( \overset{\to }{\mathop{C}}\,.\overset{\to }{\mathop{B}}\,.\overset{\to }{\mathop{A}}\,-\overset{\to }{\mathop{C}}\,.\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{B}}\, \right) \\
 & =\overset{\to }{\mathop{C}}\,.\overset{\to }{\mathop{A}}\,\overset{\to }{\mathop{B}}\,-\overset{\to }{\mathop{C}}\,.\overset{\to }{\mathop{B}}\,\overset{\to }{\mathop{A}}\, \\
\end{align}$
One can get confuse between $\overset{\to }{\mathop{A}}\,\times \left( \overset{\to }{\mathop{B}}\,\times \overset{\to }{\mathop{C}}\, \right)\And \left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\, \right)\times \overset{\to }{\mathop{C.}}\,$