The value of i^×(i^×a→)+j^×(j^×a→)+k^×(k^×a→) is :

Remember a→×(b→×c→)=b→(a→.c→)−c→(a→.b→) This is called the vector product. Using above, i^×(i^×a→)=i^.(i^.a→)−a→(i^.i^)=i^(i^.a→)−a→ j^×(j^×a→)=j^.(j^.a→)−a→(j^.j^)=j^(j^.a→)−a→ k^×(k^×a→)=k^.(k^.a→)−a→(k^.k^)=k^(k^.a→)−a→ ∴ i^×(i^×a→)+j^.(j^×a→)+k^×(k^×a→) =i^.(i^.a→)+j^.(j^.a→)+k^.(k^.a→)−3a→ ….(i) Since a→=i^ax+j^by+k^az And ax=i^.a→; ay=j^.a→; az=k^.a→ ∴ i^(i^.a→)+j^(j^.a→)+k^(k^.a→)=i^ax+j^ay+k^az=a→ On putting in eq. (i) we get i^×(i^×a→)+j^×(j^×a→)+k^×(k^×a→)=a→−3a→ =−2a→

The value of i^×(i^×a→)+j^×(j^×a→)+k^×(k^×a→) is :

Remember a→×(b→×c→)=b→(a→.c→)−c→(a→.b→) This is called the vector product. Using above, i^×(i^×a→)=i^.(i^.a→)−a→(i^.i^)=i^(i^.a→)−a→ j^×(j^×a→)=j^.(j^.a→)−a→(j^.j^)=j^(j^.a→)−a→ k^×(k^×a→)=k^.(k^.a→)−a→(k^.k^)=k^(k^.a→)−a→ ∴ i^×(i^×a→)+j^.(j^×a→)+k^×(k^×a→) =i^.(i^.a→)+j^.(j^.a→)+k^.(k^.a→)−3a→ ….(i) Since a→=i^ax+j^by+k^az And ax=i^.a→; ay=j^.a→; az=k^.a→ ∴ i^(i^.a→)+j^(j^.a→)+k^(k^.a→)=i^ax+j^ay+k^az=a→ On putting in eq. (i) we get i^×(i^×a→)+j^×(j^×a→)+k^×(k^×a→)=a→−3a→ =−2a→

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The value of $\hat{i} \times (\hat{i} \times \vec{a}) + \hat{j} \times (\hat{j} \times \vec{a}) + \hat{k} \times (\hat{k} \times \vec{a})$ is :

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$\vec{a}$

$\vec{a} \times K$

$- 2 \vec{a}$

$- \vec{a}$

answer is C.

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Detailed Solution

Remember $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} (\vec{a} . \vec{c}) - \vec{c} (\vec{a} . \vec{b})$

This is called the vector product. Using above,

$\hat{i} \times (\hat{i} \times \vec{a}) = \hat{i} . (\hat{i} . \vec{a}) - \vec{a} (\hat{i} . \hat{i}) = \hat{i} (\hat{i} . \vec{a}) - \vec{a}$

$\hat{j} \times (\hat{j} \times \vec{a}) = \hat{j} . (\hat{j} . \vec{a}) - \vec{a} (\hat{j} . \hat{j}) = \hat{j} (\hat{j} . \vec{a}) - \vec{a}$

$\hat{k} \times (\hat{k} \times \vec{a}) = \hat{k} . (\hat{k} . \vec{a}) - \vec{a} (\hat{k} . \hat{k}) = \hat{k} (\hat{k} . \vec{a}) - \vec{a}$

$∴$ $\hat{i} \times (\hat{i} \times \vec{a}) + \hat{j} . (\hat{j} \times \vec{a}) + \hat{k} \times (\hat{k} \times \vec{a})$

$= \hat{i} . (\hat{i} . \vec{a}) + \hat{j} . (\hat{j} . \vec{a}) + \hat{k} . (\hat{k} . \vec{a}) - 3 \vec{a}$ ….(i)

Since $\vec{a} = \hat{i} a_{x} + \hat{j} b_{y} + \hat{k} a_{z}$

And $a_{x} = \hat{i} . \vec{a}; a_{y} = \hat{j} . \vec{a}; a_{z} = \hat{k} . \vec{a}$

$∴$ $\hat{i} (\hat{i} . \vec{a}) + \hat{j} (\hat{j} . \vec{a}) + \hat{k} (\hat{k} . \vec{a}) = \hat{i} a_{x} + \hat{j} a_{y} + \hat{k} a_{z} = \vec{a}$