The vector i^×a.⇀b⇀i+j^×a⇀.b⇀j^+k^×a⇀.b⇀k^ is equal to :

(i×a·b)i+(j×a·b)k+(k×a·b)k=(i·(a×b))i+(j·(a×b))k+(k·(a×b)) k =(a×b)

The vector i^×a.⇀b⇀i+j^×a⇀.b⇀j^+k^×a⇀.b⇀k^ is equal to :

(i×a·b)i+(j×a·b)k+(k×a·b)k=(i·(a×b))i+(j·(a×b))k+(k·(a×b)) k =(a×b)

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$The vector (\hat{i} \times \overset{⇀}{a .} \overset{⇀}{b}) i + (\hat{j} \times \overset{⇀}{a} . \overset{⇀}{b}) \hat{j} + (\hat{k} \times \overset{⇀}{a} . \overset{⇀}{b}) \hat{k} is equal to :$

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$\overset{⇀}{b} \times \overset{⇀}{a}$

$\overset{⇀}{a}$

$\overset{⇀}{a} \times \overset{⇀}{b}$

$\overset{⇀}{b}$

answer is C.

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Detailed Solution

$\begin{array}{l} (i \times a \cdot b) i + (j \times a \cdot b) k + (k \times a \cdot b) k \\ = (i \cdot (a \times b)) i + (j \cdot (a \times b)) k + (k \cdot (a \times b)) k \end{array} = (a \times b)$

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The vector i^×a.⇀b⇀i+j^×a⇀.b⇀j^+k^×a⇀.b⇀k^ is equal to :

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$The vector (\hat{i} \times \overset{⇀}{a .} \overset{⇀}{b}) i + (\hat{j} \times \overset{⇀}{a} . \overset{⇀}{b}) \hat{j} + (\hat{k} \times \overset{⇀}{a} . \overset{⇀}{b}) \hat{k} is equal to :$