Question

The volume of parallelopipped formed by following 3 vectors will be $\underset{–}{}$cubic units.

$\overrightarrow{a}=3\hat{i}-2\hat{j}+5\hat{k}\overrightarrow{b}=2\hat{i}+2\hat{j}-\hat{k}\overrightarrow{c}=-4\hat{i}+3\hat{j}+2\hat{k}$

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Answer 1

As discussed in the hint the volume of parallelopipped spanned by the given 3 vectors is given by their scalar triple product.

So that is exactly what we will calculate.

$So[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})=\left|\begin{array}{ccc}3& -2& 5\\ 2& 2& -1\\ -4& 3& 2\end{array}\right|$

The first row contains the first vector, the second row contains the second vector and the third row contains the third vector. So above determinant will be equal to

$3\left(\right(2\times 2)-(3\times -1\left)\right)+2\left(\right(2\times 2)-(-1\times -4\left)\right)+5\left(\right(2\times 3)-(2\times 4\left)\right)$ = 21 + 70 =91 cubic units