If 4a→+5b→+9c→=0, then a→×b→×b→×c→×c→×a→ is equal to

4a+5b+9c=0 0+5(a×b)+9(a×c)=0⇒ 5(a×b)=9 (c×a) ∴ a×b, c×a Parallel(a×b)×((b×c)×(c×a))=0

If 4a→+5b→+9c→=0, then a→×b→×b→×c→×c→×a→ is equal to

4a+5b+9c=0 0+5(a×b)+9(a×c)=0⇒ 5(a×b)=9 (c×a) ∴ a×b, c×a Parallel(a×b)×((b×c)×(c×a))=0

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If $4 \vec{a} + 5 \vec{b} + 9 \vec{c} = 0, then (\vec{a} \times \vec{b}) \times [(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})]$ is equal to

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a vector perpendicular to the plane of $\vec{a}, \vec{b}$ and $\vec{c}$

a scalar quantity

$\vec{0}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} 4 a + 5 b + 9 c & = & 0 \end{array}$
$\begin{array}{rcl} \end{array}$ $\begin{array}{rcl} 0 + 5 (a \times b) + 9 (a \times c) & = & 0 \end{array}$
$\begin{array}{cl} \Rightarrow & 5 (a \times b) = 9 (c \times a) \\ ∴ a \times b, c \times a Parallel \end{array}$
$\begin{array}{rcl} (a \times b) \times ((b \times c) \times (c \times a)) & = & 0 \end{array}$