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Question

Before anyone claims that I am not studying, my university recently changed their coursework that does not require linear algebra as a pre-requisite to multi-variable calculus.

My professor gave us these questions even though we have no experience with matrix, he did not even teach us about it and I had to study them online by myself.

I only got as far as I could and I only want one question help so I can work on the other similar questions by myself. I would not be here if I can't find any other help.

Zealotory
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1 Answers1

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Figure

The volume of the parallelepiped is the base area times the height. The base area, as you know, is the magnitude of $\vec a\times\vec b$, that is $\|\vec a\times\vec b\|$.

You also know that $\vec a\times\vec b$ is a vector perpendicular to the plane containing $\vec a,\vec b$. Thus, the height of the parallelepiped is the absolute value of the projection of $\vec c$ along $\vec a\times\vec b$, that is $\Big|\vec c\cdot\frac{\vec a\times\vec b}{\|\vec a\times\vec b\|}\Big|$ (for the derivation of this quantity, see this answer).

Multiply them together to get$$V=\Big|\vec c\cdot\frac{\vec a\times\vec b}{\|\vec a\times\vec b\|}\Big|\|\vec a\times\vec b\|=|\vec c\cdot\vec a\times\vec b|$$

$\vec c\cdot\vec a\times\vec b$ is called the scalar triple product of vectors $\vec a,\vec b,\vec c$ and denoted as $[\vec a\ \vec b\ \vec c]$. The volume of the parallelepiped formed by three vectors is the geometrical interpretation of the absolute value of their scalar triple product.

In your case, note that $\vec a,\vec b,\vec c$ are coplanar, as $\vec b-\vec c=\vec a$. The volume of the parallelepiped formed by three vectors lying in the same plane is zero. You can verify the same using scalar triple product.

Shubham Johri
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