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For any vector \vec{a} , the value of (\vec{a}\times \hat{i})^{2}+(\vec{a}\times \hat{j})^{2}+(\vec{a}\times \hat{k})^{2} is equal to

  • Option 1)

    \vec{a}\, ^{2}\;

  • Option 2)

    \; 3\vec{a}\, ^{2}\;

  • Option 3)

    \; 4\vec{a}\, ^{2}\;

  • Option 4)

    \; 2\vec{a}\, ^{2}

 

Answers (1)

best_answer

As we have learned

Vector Product of Unit Vectors -

\vec{i}\times \vec{j}=\vec{k}

\vec{j}\times \vec{k}=\vec{i}

\vec{k}\times \vec{i}=\vec{j}

- wherein

\vec{i},\vec{j},\vec{k} are unit vectors

 

 \vec a = x \hat i + y \hat j + z \hat k

(\vec a\times \hat i)^2+(\vec a\times \hat j)^2+(\vec a\times \hat k)^2

(-y \hat k +z \hat j)^2 + (x\hat k -z \hat i )^2 + (-x \hat j +y \hat i )^2

= y^2 + z^2 +x^2 +z^2 +x^2 +y^2

= 2(x^2 + y^2 +z^2 )= 2|\vec a |^2

 

 

 

 

 


Option 1)

\vec{a}\, ^{2}\;

Option 2)

\; 3\vec{a}\, ^{2}\;

Option 3)

\; 4\vec{a}\, ^{2}\;

Option 4)

\; 2\vec{a}\, ^{2}

Posted by

Himanshu

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