If $\vec{a}=\vec{b}+\vec{c}, \vec{b} \times \vec{d}=\vec{0}, \vec{c} . \vec{d}=0$, then the vector $\frac{\vec{d} \times(\vec{a} \times \vec{d})}{|\vec{d}|^2}$ is always equal to:
Answer & explanation
Correct answer: option 4
$\vec{a}=\vec{b}+\vec{c}, \vec{b} \times \vec{d}=0, \vec{c} . \vec{d}=0$
$\Rightarrow \vec{a} \times \vec{d}=\vec{b} \times \vec{d}+\vec{c} \times \vec{d}=\vec{c} \times \vec{d}$
$\Rightarrow \vec{d} \times(\vec{a} \times \vec{d})=\vec{d} \times(\vec{c} \times \vec{d})$
$=(\vec{d} . \vec{d}) \vec{c}-(\vec{c} . \vec{d}) \vec{d}$
$=|\vec{d}|^2 \vec{c}$
$\Rightarrow \frac{\vec{d} \times(\vec{a} \times \vec{d})}{|\vec{d}|^2}=\vec{c}$
Hence (4) is correct answer.