The vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] are not perpendicular and \[\overrightarrow{c}\] , \[\overrightarrow{d}\] are two vectors satisfying \[\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{d}\] and \[\overrightarrow{a}.\overrightarrow{d}=0\] . The vector \[\overrightarrow{d}\] is equal to
A.\[\overrightarrow{b}-\dfrac{\overrightarrow{c}(\overrightarrow{b}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}\]

B.\[\overrightarrow{c}+\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}\]
 C.\[\overrightarrow{b}+\dfrac{\overrightarrow{c}(\overrightarrow{b}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}\]
 D.\[\overrightarrow{c}-\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}\]

Hint: We have four vectors \[\overrightarrow{a}\] , \[\overrightarrow{b}\] , \[\overrightarrow{c}\] , and \[\overrightarrow{d}\] . We have two conditions \[b\times c=b\times d\] and \[a.d=0\] . Cross multiply by \[\overrightarrow{a}\] in the LHS and RHS of \[\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{d}\] . We know the formula,
\[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})-\overrightarrow{c}(\overrightarrow{a}.\overrightarrow{b})\] . Now, using this formula, solve the equation, \[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{d})\] . Then, use the condition \[\overrightarrow{a}.\overrightarrow{d}=0\] , as given in the question. Now, divide by \[-(\overrightarrow{a}.\overrightarrow{b})\] in LHS and RHS after expanding \[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{d})\] using the formula \[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})-\overrightarrow{c}(\overrightarrow{a}.\overrightarrow{b})\] . Solve them further.

Complete step-by-step answer:
According to the question, it is given that we have four vectors \[\overrightarrow{a}\] , \[\overrightarrow{b}\] , \[\overrightarrow{c}\] , and \[\overrightarrow{d}\] . The vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] are not perpendicular. We also have two conditions given in the question.
\[\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{d}\] ……………………(1)
\[\overrightarrow{a}.\overrightarrow{d}=0\] ………………………….(2)
Now, multiplying by \[\overrightarrow{a}\] in equation (1), we get
\[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{d})\] ………………………………..(3)
We need to simplify equation (3) and we also know the formula, \[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})-\overrightarrow{c}(\overrightarrow{a}.\overrightarrow{b})\] .
Using this formula, simplifying equation (3)
\[\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{d})\]
\[\Rightarrow \overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})-\overrightarrow{c}(\overrightarrow{a}.\overrightarrow{b})=\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{d})-\overrightarrow{d}(\overrightarrow{a}.\overrightarrow{b})\] ………………………..(4)
Now, dividing by \[-(\overrightarrow{a}.\overrightarrow{b})\] in equation (4), we get
\[\Rightarrow \dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{-(\overrightarrow{a}.\overrightarrow{b})}-\dfrac{\overrightarrow{c}(\overrightarrow{a}.\overrightarrow{b})}{-(\overrightarrow{a}.\overrightarrow{b})}=\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{d})}{-(\overrightarrow{a}.\overrightarrow{b})}-\dfrac{\overrightarrow{d}(\overrightarrow{a}.\overrightarrow{b})}{-(\overrightarrow{a}.\overrightarrow{b})}\]
\[\Rightarrow -\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}+\overrightarrow{c}=\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{d})}{-(\overrightarrow{a}.\overrightarrow{b})}+\overrightarrow{d}\] …………………………(5)
From equation (2), we have \[\overrightarrow{a}.\overrightarrow{d}=0\] .
Now, putting \[\overrightarrow{a}.\overrightarrow{d}=0\] in equation (5), we get
 \[\Rightarrow -\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}+\overrightarrow{c}=\dfrac{\overrightarrow{b}(0)}{-(\overrightarrow{a}.\overrightarrow{b})}+\overrightarrow{d}\]
\[\begin{align}
  & \Rightarrow \overrightarrow{c}-\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}=0+\overrightarrow{d} \\
 & \Rightarrow \overrightarrow{c}-\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}=\overrightarrow{d} \\
\end{align}\]
So, \[\overrightarrow{d}=\overrightarrow{c}-\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}\] .
Hence, the option (D) is the correct option.

Note: In this question, one may think to do cross-multiplication by \[\overrightarrow{b}\] in the LHS and RHS of the equation \[\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{d}\] . If we do so, then we will not be able to use the condition \[\overrightarrow{a}.\overrightarrow{d}=0\] .
\[\overrightarrow{b}\times (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{b}\times (\overrightarrow{b}\times \overrightarrow{d})\]
\[\Rightarrow \overrightarrow{b}(\overrightarrow{b}.\overrightarrow{c})-\overrightarrow{c}(\overrightarrow{b}.\overrightarrow{b})=\overrightarrow{b}(\overrightarrow{b}.\overrightarrow{d})-\overrightarrow{d}(\overrightarrow{b}.\overrightarrow{b})\]
Also, in the question conditions for \[\overrightarrow{b}.\overrightarrow{d}\] and \[\overrightarrow{b}.\overrightarrow{c}\]is not given. So, we can’t approach this question by this method.