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Question: The vector \[\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)+\widehat{j}\times...

The vector i^×(a→×i^)+j^×(a→×j^)+k^×(a→×k^)\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)+\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right)+\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right) is equal to
(a) 0→\overrightarrow{0}
(b) a→\overrightarrow{a}
(c) 2a→2\overrightarrow{a}
(d) None of these

Explanation

Solution

First of all, assume, a→=xi^+yj^+zk^\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}. Now use the formula for vector triple product that is a→×(b→×c→),(a→.c→)b−(a→.b→)c\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right),\left( \overrightarrow{a}.\overrightarrow{c} \right)b-\left( \overrightarrow{a}.\overrightarrow{b} \right)c and also use i^.i^=j^.j^=k^.k^=1;i^.j^=j^.k^=k^.i^=0\widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1;\widehat{i}.\widehat{j}=\widehat{j}.\widehat{k}=\widehat{k}.\widehat{i}=0 to find the value of the given expression. Also separate the given expression and then solve individually each pat to avoid any confusion.

Complete step-by-step solution -
In this question, we have to find the value of the vector
i^×(a→×i^)+j^×(a→×j^)+k^×(a→×k^)\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)+\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right)+\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right)
First of all, let us consider the expression given in the question.
E=i^×(a→×i^)+j^×(a→×j^)+k^×(a→×k^)E=\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)+\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right)+\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right)
Let us take the value of a→=xi^+yj^+zk^\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}.
Let us assume,

& \overrightarrow{P}=\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right) \\\ & \overrightarrow{Q}=\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right) \\\ & \overrightarrow{R}=\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right) \\\ \end{aligned}$$ So, we get the above expression as, $$E=\overrightarrow{P}+\overrightarrow{Q}+\overrightarrow{R}....\left( i \right)$$ We know that according to the vector triple product, $$\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}....\left( ii \right)$$ By using this, let us find the value of vectors P, Q, and R. Let us find the value of vector P. $$\overrightarrow{P}=\widehat{i}\times \left( \overrightarrow{a}\times \widehat{i} \right)$$ By substituting the value of $$\overrightarrow{a}=\widehat{i},\overrightarrow{b}=\overrightarrow{a},\overrightarrow{c}=\widehat{i}$$in equation (ii), we get, $$\overrightarrow{P}=\left( \widehat{i}.\widehat{i} \right)\overrightarrow{a}-\left( \widehat{i}.\overrightarrow{a} \right)\widehat{i}$$ By substituting, $$\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}$$, we get, $$\overrightarrow{P}=\left( \widehat{i}.\widehat{i} \right)\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ \widehat{i}.\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right) \right]\widehat{i}$$ We know that, $$\begin{aligned} & \left( a\widehat{i}+b\widehat{j}+c\widehat{k} \right).\left( p\widehat{i}+q\widehat{j}+r\widehat{k} \right)=ap+bq+cr \\\ & \widehat{i}.\widehat{i}=\widehat{j}.\widehat{j}=\widehat{k}.\widehat{k}=1 \\\ & \widehat{i}.\widehat{j}=\widehat{j}.\widehat{k}=\widehat{k}.\widehat{i}=0 \\\ \end{aligned}$$ By using these, we get, $$\overrightarrow{P}=1\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ x+0+0 \right]\widehat{i}$$ $$\overrightarrow{P}=x\widehat{i}+y\widehat{j}+z\widehat{k}-x\widehat{i}$$ $$\overrightarrow{P}=y\widehat{j}+z\widehat{k}....\left( iii \right)$$ Now, let us find the value of vector Q. $$\overrightarrow{Q}=\widehat{j}\times \left( \overrightarrow{a}\times \widehat{j} \right)$$ By substituting the value of $$\overrightarrow{a}=\widehat{j},\overrightarrow{b}=\overrightarrow{a},\overrightarrow{c}=\widehat{j}$$in equation (ii), we get, $$\overrightarrow{Q}=\left( \widehat{j}.\widehat{j} \right)\overrightarrow{a}-\left( \widehat{j}.\overrightarrow{a} \right)\widehat{j}$$ By substituting, $$\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}$$, we get, $$\overrightarrow{Q}=\left( \widehat{j}.\widehat{j} \right)\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ \widehat{j}.\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right) \right]\widehat{j}$$ $$\overrightarrow{Q}=1\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ 0+y+0 \right]\widehat{j}$$ $$\overrightarrow{Q}=x\widehat{i}+y\widehat{j}+z\widehat{k}-y\widehat{j}$$ $$\overrightarrow{Q}=x\widehat{i}+z\widehat{k}....\left( iv \right)$$ Now, let us find the value of vector R. $$\overrightarrow{R}=\widehat{k}\times \left( \overrightarrow{a}\times \widehat{k} \right)$$ By substituting the value of $$\overrightarrow{a}=\widehat{k},\overrightarrow{b}=\overrightarrow{a},\overrightarrow{c}=\widehat{k}$$in equation (ii), we get, $$\overrightarrow{R}=\left( \widehat{k}.\widehat{k} \right)\overrightarrow{a}-\left( \widehat{k}.\overrightarrow{a} \right)\widehat{k}$$ By substituting, $$\overrightarrow{a}=x\widehat{i}+y\widehat{j}+z\widehat{k}$$, we get, $$\overrightarrow{R}=\left( \widehat{k}.\widehat{k} \right)\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ \widehat{k}.\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right) \right]\widehat{k}$$ $$\overrightarrow{R}=1\left[ x\widehat{i}+y\widehat{j}+z\widehat{k} \right]-\left[ 0+0+z \right]\widehat{k}$$ $$\overrightarrow{R}=x\widehat{i}+y\widehat{j}+z\widehat{k}-z\widehat{k}$$ $$\overrightarrow{R}=x\widehat{i}+y\widehat{j}....\left( v \right)$$ Now, by substituting the values of vectors P, Q and R from equation (iii), (iv) and (v), we get, $$E=y\widehat{j}+z\widehat{k}+x\widehat{i}+z\widehat{k}+x\widehat{i}+y\widehat{j}$$ $$E=2x\widehat{i}+2y\widehat{j}+2z\widehat{k}$$ $$E=2\left( x\widehat{i}+y\widehat{j}+z\widehat{k} \right)$$ By replacing $$x\widehat{i}+y\widehat{j}+z\widehat{k}=\overrightarrow{a}$$, we get, $$E=2\overrightarrow{a}$$ Hence, we get the value of the given expression as $$2\overrightarrow{a}$$. Hence, the option (c) is the right answer.

Note: In these types of questions, whenever the value of a certain vector is not given, we should always consider it as a general vector that is xi^+yj^+zk^x\widehat{i}+y\widehat{j}+z\widehat{k}. Also, note that the vector triple product of 3 vectors is not associative that is, (a→×b→)×c→≠a→×(b→×c→)\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}\ne \overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right) because

& \left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a} \\\ & \overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c} \\\ \end{aligned}$$ So, it is advisable to apply the formula carefully to avoid any mistakes.