# Tag Info

1

Yes you need a vector which is perpendicular to the normal $(1, 1, 1)$ of the plane and perpendicular to $(0, 1, -1)$, so you need to work out the cross product

0

If the context doesn't tell you anything and you don't have any parentheses you should evaluate from left to right as long as I learned.

1

Since cross product is not associative $A\times B\times C$ is meaningless.

1

Choose coordinates so that the two vectors $\vec a, \vec b$ are in the $xy$-plane, with $\vec a$ along the $x$-axis. (Note that as long as you've decided on a unit length, exactly which direction you choose for the coordinate axes doesn't change anything. The vectors and their cross product lives in a coordinate-free space, just floating around. We're just ...

1

It is probably because the answer is simple in terms of classical 2D geometry. $||\vec u\times \vec v||=||\vec u||.||\vec v||.\sin(\vec u,\vec v)$ But the area of the parallelogram defined by $\vec u$ and $\vec v$ is the base multiplied by the height. If you take $\vec u$ as the base, the height is $h=||\vec v||.\sin(\vec u,\vec v)$, hence the result...

2

The volume of the parallelepiped is given by $\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})$. The LHS of your expression is the triple scalar product using the vectors $\mathbf{a}\times\mathbf{b}$, $\mathbf{b}\times\mathbf{c}$ and $\mathbf{c}\times\mathbf{a}$. These vectors represent vectors that are orthogonal to the corresponding faces of the parallelepiped, ...

2

$u \times v$ can be defined as the vector that makes $$\langle w, u \times v \rangle=\det(w,u,v) \quad \forall w.$$ That there is such a vector can be seen by evaluating the above equality with the canonical basis. That it is unique can be seen in the same way. Now your result follows easily.

1

$u\cdot(v\times w)$ is the determinant of the matrix whose columns are $u$, $v$ and $w$. (This can be used as a definition for the cross product). The identity now follows from the usual determinant rules for permuting rows/columns.

4

$$u \cdot (v \times w) = \det(u, v, w)$$ where (u, v, w) means the 3x3 matrix whose columns are $u$, $v$, and $w$ respectively, and thus the triple product satisfies all of the algebraic identities that determinants do. (alternatively, $(u,v,w)$ can mean the matrix whose rows are $u$, $v$, and $w$, if you prefer to think of vectors as row vectors)

2

Yes. The volume of the parallelepipid is equal to the scalar triple product. Consider the geometric definitions of the cross-product and inner-product to see this.

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