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Question: Let \( \overrightarrow a = \widehat j - \widehat k\;{\text{and}}\;\overrightarrow c = \widehat i - \...

Let a=j^k^  and  c=i^j^k^.\overrightarrow a = \widehat j - \widehat k\;{\text{and}}\;\overrightarrow c = \widehat i - \widehat j - \widehat k. Then the vector satisfying a×b+c=0  anda.b=3\overrightarrow a \times \overrightarrow b + \overrightarrow c = \overrightarrow 0 \;{\text{and}}\,\overrightarrow a .\overrightarrow b = 3 is
A. i^+j^2k^- \widehat i + \widehat j - 2\widehat k
B. 2i^j^+2k^2\widehat i - \widehat j + 2\widehat k
C. i^j^2k^\widehat i - \widehat j - 2\widehat k
D. i^+j^2k^\widehat i + \widehat j - 2\widehat k

Explanation

Solution

Hint : To find the value of b\overrightarrow b you have to remember the formula for vector triple product, here in the first given equation do cross product with a\overrightarrow a both sides and then use the vector triple product formula and the given information in the question to elaborate and find the value of b\overrightarrow b
Formula used: Vector triple product: a×b×c=(a.c)b(a.b)c\overrightarrow a \times \overrightarrow b \times \overrightarrow c = \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow c

Complete step-by-step answer :
In order to find b\overrightarrow b we will see what we have given in the question,
So we have values of vector a  and  c\overrightarrow a \;{\text{and}}\;\overrightarrow c and also two relations or you may say vector equations that are
a×b+c=0  anda.b=3\overrightarrow a \times \overrightarrow b + \overrightarrow c = \overrightarrow 0 \;{\text{and}}\,\overrightarrow a .\overrightarrow b = 3
Now, in the first equation, cross multiplying a\overrightarrow a both sides, we will get
a×(a×b+c)=a×0 a×(a×b)+a×c=0  \Rightarrow \overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b + \overrightarrow c } \right) = \overrightarrow a \times \overrightarrow 0 \\\ \Rightarrow \overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow b } \right) + \overrightarrow a \times \overrightarrow c = \overrightarrow 0 \\\
Using the vector triple product formula to further evaluate the equation, we will get
(a.b)a(a.a)b+a×c=0\Rightarrow \left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow a - \left( {\overrightarrow a .\overrightarrow a } \right)\overrightarrow b + \overrightarrow a \times \overrightarrow c = \overrightarrow 0
Evaluating the equation furthermore by putting value of a.b\overrightarrow a .\overrightarrow b from the question and simplifying the cross product a×c\overrightarrow a \times \overrightarrow c , we will get

3a((j^k^).(j^k^))b+(j^k^)×(i^j^k^)=0 3a2b+(k^i^j^i^)=0 3a2b2i^k^j^=0 2b=3a2i^k^j^ 2b=3(j^k^)2i^k^j^ 2b=3j^3k^2i^k^j^ b=2i^+2j^4k^2 b=i^+j^2k^   \Rightarrow 3\overrightarrow a - \left( {\left( {\widehat j - \widehat k} \right).\left( {\widehat j - \widehat k} \right)} \right)\overrightarrow b + \left( {\widehat j - \widehat k} \right) \times \left( {\widehat i - \widehat j - \widehat k} \right) = \overrightarrow 0 \\\ \Rightarrow 3\overrightarrow a - 2\overrightarrow b + \left( { - \widehat k - \widehat i - \widehat j - \widehat i} \right) = \overrightarrow 0 \\\ \Rightarrow 3\overrightarrow a - 2\overrightarrow b - 2\widehat i - \widehat k - \widehat j = \overrightarrow 0 \\\ \Rightarrow 2\overrightarrow b = 3\overrightarrow a - 2\widehat i - \widehat k - \widehat j \\\ \Rightarrow 2\overrightarrow b = 3\left( {\widehat j - \widehat k} \right) - 2\widehat i - \widehat k - \widehat j \\\ \Rightarrow 2\overrightarrow b = 3\widehat j - 3\widehat k - 2\widehat i - \widehat k - \widehat j \\\ \Rightarrow \overrightarrow b = \dfrac{{ - 2\widehat i + 2\widehat j - 4\widehat k}}{2} \\\ \Rightarrow \overrightarrow b = - \widehat i + \widehat j - 2\widehat k \;

Therefore b=i^+j^2k^\overrightarrow b = - \widehat i + \widehat j - 2\widehat k is the required value.
So, the correct answer is “Option B”.

Note : Normally multiplication holds good for commutative property but in the case of vector products or vector triple product commutative property does not hold true, so if you are getting your answers wrong in solving vector products then may check this once. In simple words, a×b×ca×c×b\overrightarrow a \times \overrightarrow b \times \overrightarrow c \ne \overrightarrow a \times \overrightarrow c \times \overrightarrow b
This question can be solved in one more way, in which we have to initially consider the value of vector bb to be mixture of some variables and then perform the given equation or operation, after performing the operation, you have to compare each terms with their respective terms in the right hand side to get the values of variables and eventually get the value of vector bb