# Tag Info

0

So it is understood that you are looking for the value of $s$ in $$s = \vec u \cdot \left( {\vec v \times \vec w} \right)$$ First note that $s$ is a scalar, that is a real number (the volume - with sign - of the paralleliped defined by the three vectors). Second, since ${\vec v}$ and ${\vec w}$ are orthogonal to each other, their cross product will be a ...

2

Hint: Note that $$u \cdot (v \times w) = w \cdot (u \times v)$$ That is, we can cyclically permute the vectors in a triple scalar product.

1

Consider $p \times q = 3p \times r \implies p \times q = p \times 3r$ Now , $p \times q - p \times 3r = 0 \implies p \times (q-3r) = 0$ Now if cross product of two vectors is 0 , one must a multiple of other so that cross-product of the same vectors is 0. So , $q-3r = \lambda p$ which implies on substituting $p \times \lambda p = \lambda p \times p ... 0 The normal to a surface of the form$f=c$is just the gradient of$f$, in your example $$\nabla f = (2x, -2y, -2z)^T =\left( \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)^T$$ whenever$(x,y,z)$fulfils your equation. This is assuming the gradient does not vanish (in which case the equation does not define a ... 0 Given a surface of equation $$z=f(x,y)$$ you can find the normal vector$N$at a point$P(x_0,y_0)$as: $$N=[f_x(x_0,y_0),f_y(x_0,y_0),-1]^T$$ where:$f_x=\dfrac{\partial f}{\partial x}$and$f_y=\dfrac{\partial f}{\partial y}$If the equation has given in implicit form you can get for$N$: $$N=[f_x(x_0,y_0,z_0),f_y(x_0,y_0,z_0),f_z(x_0,y_0,z_0)]^T$$ 0 Basically we want a coordinate-free proof that if$\mathbf{n}$is a unit normal vector of an oriented plane$\Pi$then$(\mathbf{a}\times\mathbf{b})\cdot\mathbf{n}$is the signed area of the parallelogram spanned by the orthogonal projections of the two vectors$\mathbf{a}$and$\mathbf{b}$onto$\Pi$. (My definition of$\mathbf{a}\times\mathbf{b}$is the ... 0 There is indeed a geometric interpretation of$\bf{u}\times\bf{v}$in terms of the areas of the projections of the parallelogram$\bf{P}$spanned by$\bf{v}$and$\bf{w}$onto the coordinate planes. I'll start from scratch. Motivating problem: We wish to create a vector perpendicular to u,v, i.e. construct w s.t.$\bf{w}\cdot \bf{u} = \bf{w}\cdot \bf{...

3

There is another universal property that is related and much more well known, which is the universal property of the exterior algebra. In particular, there is a universal property for each $\Lambda^k(V)$, and the one we need is the following: Universal Property of the Exterior Product: Let $V$ be a vector space. Given any alternating bilinear map $\mu: V\... 3 Here is how I usually introduce the cross product. Given two vectors$u=(u_1,u_2,u_3), v=(v_1,v_2,v_3)$we seek another vector$w=(x,y,z)$which is perpendicular to both. This means $$u \cdot w= 0 \Rightarrow u_1x+u_2y+u_3z=0 \\ v \cdot w =0 \Rightarrow v_1x+v_2y+v_3z=0$$ Now, all you have to do is solve this system of equations.Under the extra ... 3 To me, the wedge product is more intuitive than the cross product. And as it turns out, geometric algebra provides exactly the way to get a vector perpendicular to the bivector$u\wedge v$: taking the algebraic dual. So I define the cross product as $$\bbox[5px,border:2px solid red]{u\times v := (u\wedge v)I^{-1}}$$ where$I$is the positively oriented ... 2 I don't know what you mean by derive here, but by definition the two products give different types of geometric objects. The exterior product of two vectors (in$\mathbf{R}^n$, for any$n$) is a bivector, whereas the cross product of two vectors (in$\mathbf{R}^3$only) is another vector. For$n=3$, there happens to be an isomorphism between bivectors and ... 5 I believe the exterior products (wedge products) of two vectors$u$and$v$are denoted by$u\wedge v$. Note: the$LaTeX$command for wedge products is also \wedge. I have borrowed this picture from Wikipedia's page on exterior algebra as it explains the difference very clearly.$a\times b$is the cross product of the two vectors$a$and$b\$. The cross ...

Top 50 recent answers are included