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Question: If \(\mathbf{a} \times (\mathbf{b} + \mathbf{c}) + \mathbf{b} \times (\mathbf{c} + \mathbf{a}) + \ma...

If a×(b+c)+b×(c+a)+c×(a+b)=\mathbf{a} \times (\mathbf{b} + \mathbf{c}) + \mathbf{b} \times (\mathbf{c} + \mathbf{a}) + \mathbf{c} \times (\mathbf{a} + \mathbf{b}) =, c=(2,−2,4)\mathbf{c} = (2, - 2,4) and i is the unit vector in the x-direction, then (a−2b+3c).i=(a - 2b + 3c).i =

A

11

B

15

C

18

D

36

Answer

11

Explanation

Solution

a=(1,−1,2),b=(−2,3,5),c=(2,−2,4)\mathbf{a} = (1, - 1,2),\mathbf{b} = ( - 2,3,5),\mathbf{c} = (2, - 2,4)

So, a=(1,−1,2)≡i−j+2k;b=(−2,3,5)≡−2i+3j+5k\mathbf{a} = (1, - 1,2) \equiv \mathbf{i} - \mathbf{j} + 2\mathbf{k};\mathbf{b} = ( - 2,3,5) \equiv - 2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}

and c=(2,−2,4)≡2i−2j+4k\mathbf{c} = (2, - 2,4) \equiv 2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}

⇒a−2b+3c=(i−j+2k)−2(−2i+3j+5k)+3(2i−2j+4k)\Rightarrow \mathbf{a} - 2\mathbf{b} + 3\mathbf{c} = (\mathbf{i} - \mathbf{j} + 2\mathbf{k}) - 2( - 2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) + 3(2\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})

=11i−13j+4k= 11\mathbf{i} - 13\mathbf{j} + 4\mathbf{k} and (a−2b+3c).i=11.(\mathbf{a} - 2\mathbf{b} + 3\mathbf{c}).\mathbf{i} = 11.