Problem on finding the volume of a parallelepiped
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Sketch and find the volume of the parallelepiped spanned by the vectors $\bfi, \bfi+\bfj,$ and $\bfi + \bfk.$
Solution
We sketch the three vectors:
The parallelepiped spanned by them is:
Recall that
Triple product and the volume of parallelepipeds
The volume of the parallelepiped spanned by the vectors $\bfa, \bfb, \bfc$ is $\left| \bfa \cdot (\bfb \times \bfc) \right|.$
It doesn't matter which vector is $\bfa, \bfb$ or $\bfc$, so we choose them so that the cross product is easier to compute.
Let $\bfa = \bfi + \bfj$, and $\bfb = \bfi$, and $\bfc = \bfi + \bfk.$
(see details)
We can compute that $\bfb \times \bfc = - \bfj.$
Direct computation of dot product
We can now compute the dot product $$\bfa \cdot (\bfb \times \bfc) = (\bfi + \bfj) \cdot (-\bfj) = -1, $$ which has absolute value $1$.
The volume of the parallelepiped is 1.
Related topics
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Cross product
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The volume of the parallelepiped spanned by the vectors $\bfa, \bfb, \bfc$ is $\left| \bfa \cdot (\bfb \times \bfc) \right|.$
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