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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 ...


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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.


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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 ...


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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 ...


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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 $N$: $$N=[f_x(x_0,y_0,z_0),f_y(x_0,y_0,z_0),f_z(x_0,y_0,z_0)]^T$$


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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 ...


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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{...


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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\...


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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 ...


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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 ...


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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 ...


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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 ...



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