Question

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.$\overrightarrow{c}-\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}$

Answer 1

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
$\Rightarrow -\dfrac{\overrightarrow{b}(\overrightarrow{a}.\overrightarrow{c})}{(\overrightarrow{a}.\overrightarrow{b})}+\overrightarrow{c}=\dfrac{\overrightarrow{b}(0)}{-(\overrightarrow{a}.\overrightarrow{b})}+\overrightarrow{d}$

& \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{aligned}$$ So, $$\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.