If a→=b→+c→,b→×d→=0→,c→.d→=0, then the vector d→×a→×d→d→2 is always equal to

a=b+c b×d=0 c·d=0 d×(a×d)=(d·d)a-(d·a)d a=b+c ⇒ a·d=b·d+0 ⇒ a·d=b·d d2a-(d·b)dd2=a

If a→=b→+c→,b→×d→=0→,c→.d→=0, then the vector d→×a→×d→d→2 is always equal to

a=b+c b×d=0 c·d=0 d×(a×d)=(d·d)a-(d·a)d a=b+c ⇒ a·d=b·d+0 ⇒ a·d=b·d d2a-(d·b)dd2=a

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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

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$\vec{a}$

$\vec{d}$

$\vec{b}$

$\vec{c}$

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$a = b + c b \times d = 0 c \cdot d = 0 d \times (a \times d) = (d \cdot d) a - (d \cdot a) d a = b + c \Rightarrow a \cdot d = b \cdot d + 0 \Rightarrow a \cdot d = b \cdot d |\frac{d^{2} a - (d \cdot b) d}{d^{2}}| = a$

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If a→=b→+c→,b→×d→=0→,c→.d→=0, then the vector d→×a→×d→d→2 is always equal to

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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