Question

Find the vector, not with determinants, but by using properties of cross products. k × (i − 3j)

Answer 1

$\mathbf k\times(\mathbf i-3\mathbf j)=(\mathbf k\times\mathbf i)-3(\mathbf k\times\mathbf j)$
$=\mathbf j-3(-\mathbf i)$
$=3\mathbf i+\mathbf j$

Answer 2

The vector will be "$3 \hat{i}+\hat{j}$".

According to the question,

→ $\vec{a} = \hat{k}\times (\hat{i}-3 \hat{j})$

The cross product is distributive,

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

     $=(\hat {k}\times \hat{i})+(-3) (\hat {k}\times \hat{j})$...(equation 1)

As we know,

→ $\vec{c}\times \vec{d} = -(\vec{d}\times \vec{c})$

then,

→ $\hat{k}\times \hat{j} = -(\hat {j}\times \hat{k})$

but,

→ $\hat{j}\times \hat{k} =\hat{i}$

hence,

→ $\hat{k}\times \hat{j} = -\hat{i}$...(equation 2)

also we know that,

→ $\hat{k}\times \hat{i}=\hat {j}$...(equation 3)

Now,

By substituting "$\hat {k}\times \hat{j} = -\hat{i}$" and "$\hat {k}\times \hat{i}=\hat{j}$" in the (equation 1), we get

→ $\vec{a} = \hat {j}+(-3)(-\hat{i})$

     $= \hat{j}+3\hat {i}$

     $= 3\hat {i}+\hat{j}$

Thus the above response is right.

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