# Tag Info

2

$f(x,y,z) = x \times (y \times z) + y \times (z \times x) + z \times (x \times y)$ is linear in each of $x$, $y$, $z$, so it suffices to prove for the cases where each of $x$, $y$ and $z$ is one of the three standard unit vectors $i$, $j$, $k$. Now $f(x,y,z)$ is multiplied by $-1$ if you interchange any two of $x$, $y$ and $z$: in particular if two are ...

4

There is a property which should make this question easier: $$a \times (b \times c) = b(a \cdot c) - c(a \cdot b)$$ You will notice that all the terms cancel out, leading to the answer of $0$ as desired.

0

v1(1,t,-2); v2(-3,1,6); 3*v1 is parallel to v2 if t=-1/3 Also: Perform vector Cross Product v1Xv2=v3 to get components: v3((6*t+2),(6-6),(1+3*y)); the cross product is 0 when t=-1/3

2

You can use the fact that for two vectors $u,v$, $$| u \times v | = |u| \ |v| \ \sin\theta,$$ where $\theta$ is the angle between $u$ and $v$. Since you want $u$ and $v$ to be parallel, you want $\sin\theta = 0$, so $|u\times v|=0$. This means you can solve your problem by finding the cross product and then setting its magnitude equal to $0$ and solving ...

0

I hope you know that $u\times v=-v\times u$, so cross multiplication is not commutative. So $AB \times AC$ and $AC \times AB$ will give you opposite normal vectors. However, this should not affect your later work: any normal vector will do for this problem. So your intermediate calculations will change if you use the other cross product, but the end result ...

4

Because the cross product is not defined merely for the one purpose of writing rotations efficiently but has lots of other uses, and for many of these its algebraic properties are very useful. $a\times b$ as defined is an antisymmetric bilinear product; the quantity you define isn't linear in $a$. The cross product can only be interpreted the way you're ...

0

The angular velocity $\;\vec \omega\;$ of a particle at positional displacement $\;\vec r\;$ from the POV moving at (instantaneous) linear velocity $\;\vec v\;$ relative to the POV is indeed calculated as: $$\vec \omega = \frac {\vec r\times \vec v}{\lvert \vec r\rvert^2}$$ (NB: The Point Of View is typical taken as the centre of mass of the system, for ...

1

I don't think the dot product is derived from the cosine rule, but you can deduce the cosine rule from the dot product. The dot product formula $\mathbf{a} \cdot \mathbf{b} =\lVert \mathbf{a}\rVert \cdot\lVert \mathbf{b}\rVert \cos\theta$ should be taken as a definition of dot product. The rationale behind or purpose for the dot product is to be able to ...

0

Setting $z=0$ for every point would project every point onto the $xy$-plane. You could also choose $x=0$ to project onto the $yz$-plane, or $y=0$ to project onto the $xz$-plane, or set any other fixed linear combination of the coordinates to zero to project onto another plane, whose orientation depends on the choice of the coefficients of that linear ...

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I think the following picture can explain your doubt.

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For part #1: Let's call the two given planes $A$ and $B$, and the desired plane $C$. If $C$ is perpendicular $A$ and $B$, then the normal $N_C$ of $C$ is perpendicular to the normals $N_A$ and $N_B$ of $A$ and $B$. Draw a picture to convince yourself that this is true. So $N_C$ is parallel to the cross product $N_A \times N_B$. In your case, $A$ is the ...

1

In the first part you need the cross product of the two plane normals to find the required normal to the plane, so that you can write down the equation In the second part, find the vector joining the two points and do the cross product of this and the normal to the given plane,cso you can write down the equation in a similar way as before.

0

Here is what you want to think: Find the line through $(3,0,-1)$ that is orthogonal to each plane (use projection maps). Now you have two lines through a common point, which determines a plane. Think of the three coordinate planes in $\mathbb{R}^{3}$: no two are parallel, but each is perpendicular to both of the others. (this is just part a). For part ...

1

Alternatively, if you don't want to fuss over expanding the determinants as I told you to try, you can use this argument. Since $V:=u_1\cdot\left(u_2\times u_3\right)=u_2\cdot\left(u_3\times u_1\right)=u_3\cdot\left(u_1\times u_2\right) \neq 0$, the vectors $u_1$, $u_2$, and $u_3$ are linearly independent. Therefore, the solution $r\in\mathbb{R}^3$ to the ...

1

Let the angle between $u$ & $v$ be $\alpha$, now the dot product is given as follows $$u\cdot v=|u||v|\cos \alpha$$ $$\implies \cos \alpha=\frac{u\cdot v}{|u||v|}$$ $$=\frac{(1, -2, 3)\cdot (-4, 5, 6)}{|(1, -2, 3)||(-4, 5, 6)|}$$ $$\cos \alpha=\frac{-4-10+18}{\sqrt{(1)^2+(-2)^2+(3)^2}\sqrt{(-4)^2+(5)^2+(6)^2}}$$ $$\cos \alpha=\frac{4}{\sqrt{14\times ... 0 Hint The cross product satisfies$$||{\bf a} \times {\bf b}|| = ||{\bf a}|| \, ||{\bf b}|| \sin \theta, where $\theta \in [0, \pi]$ is the angle between $\bf a$ and $\bf b$. (In fact this property is very nearly one of the common definitions of the cross product; see, e.g., Defn. 7.4 of Dennis G. Zill, Michael R. Cullen (2006). Advanced engineering ...

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