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0

Let $P=(a_1,b_1),P_0=(a_2,b_2),\overrightarrow{d}=(a_3,b_3)$. Now vector $\overrightarrow{P_0P}=(a_2-a_1,b_2-b_1)$. We only need a length, so we can write it as $(a_1-a_2,b_1-b_2)$. Now apply formula for vector cross product ...


0

This explains the triple vector product very well: https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/vector-triple-product-expansion-very-optional


1

I'm going to use the basic definitions of scalar and vector triple products to prove this. Let $\mathbf u= a_1 \mathbf i + a_2 \mathbf j + a_3 \mathbf k, \mathbf v= b_1 \mathbf i + b_2 \mathbf j + b_3 \mathbf k, \mathbf w= c_1 \mathbf i + c_2 \mathbf j + c_3 \mathbf k$, Then $(\mathbf{v} \times \mathbf{w})= (b_2c_3-b_3c_2) \mathbf i + (b_3c_1-b_1c_3) ...


0

since the cross product is non-associative your first error is to write the ambiguous expression $P \times Q \times R$. since the scalar product gives a scalar, your second error is to write the incorrect form $(P.Q).R$ however your method of using co-ordinates should lead to the correct answer. the $i$-component of $P \times (Q \times R)$ is: $$ ...


0

Following this formula the ratio can be written as $$\sqrt\frac{\det((MA)^TMA)}{\det(A^TA)}=\sqrt\frac{\det (A^T(M^TM)A)}{\det(A^TA)}$$ where $A=\pmatrix{a & b}$.


3

Chappers' proof is the way that was most likely intended (using the identity given). However, here is another approach (if only to show how much work is saved by using the identity given) Since $(a\times b)\times a$ is perpendicular to $a$, $$ ((a\times b)\times a)\cdot a=0\tag{1} $$ Since $|a\times b|=|a|\,|b|\,|\sin(\theta)|$ and $a\cdot ...


3

If you know the identity $$x \cdot (y \times z) = (x \times y) \cdot z$$ for arbitrary vectors $x,y,z$, then putting $x=a \times b$ and $y=c$, $z=d$, you have $$ (a \times b) \cdot (c \times d) = ((a \times b) \times c) \cdot d $$ so you can arrange your second equation so that everything is dotted with $d$: $$ ((a \times b) \times c) \cdot d = ((a \cdot ...


1

$\vec k_i \cdot \vec v_j = \frac{(\vec v_j \times \vec v_k) \cdot (\vec v_j)}{\vec v_1 \cdot (\vec v_2 \times \vec v_3 )} =0 $ since the numerator contains the inner product of $\vec v_j$ with a vector orthogonal to $\vec v_j$. We may replace $j$ with $k$ to complete the proof. For ii) we have $$\begin{align} \vec k_1 \cdot (\vec k_2 \times\vec ...


1

Here is a clean way to solve this problem without any brute force, although it might involve more linear algebra than you've seen. Combine the following facts: The triple scalar product $v_1 \cdot (v_2 \times v_3)$ is the determinant of the $3 \times 3$ matrix $V$ with columns $v_1, v_2, v_3$. It is also the determinant of the $3 \times 3$ matrix with rows ...


0

Parallel vectors have a constant minimum distance between them in Euclidean geometry, a single minimum distance between them in skewed orientation of Hyperbolic geometry, a single maximum distance between them in intersecting Elliptic geometry.


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"Magnitude", "direction" make sense only if your vector space has an inner product. Cross product only makes sense, if your vector space is 3-dimensional and has an inner product. Geometrically speaking, two vectors are parallel, if they have the same "direction" (or opposite direction), regardless of "magnitudes". This construction makes sense ...


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Two vectors do not need to have the same magnitude to be parallel. Intuitively, two vectors are parallel if, when you place them on top of eachother, they form one single line. Meaning, they can have the same direction or opposite direction. This also means that if they are not on top of eachother, they will never intersect. Be aware here, that vectors are ...


0

You got tons of answers but Sami Ben Romdhane's post is worth expanding upon, in my opinion, for if you've just started studying linear algebra, this approach will be relatively new. I don't know how familiar you are with terms, so I will assume you know what $\mathbb{R}^3$ is and what matrices and determinants are, and of course the standard scalar and ...


0

Let's say you want to take the cross-product of the two vectors shown coordinate-wise below: $$v_1 \times v_2 = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} \times \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}$$ If you know about the determinant of a $3\times3$ matrix, let $x, y, z$ be unknown and write $$\Delta = \begin{vmatrix} x & y & z \\ x_1 & ...


1

The best way to intuitively understand the vector product is by thinking of it in terms of an area form. By this I mean something to which I can give a patch of a plane, and it will tell me the area. This is important, for example, to define area/surface/double integrals. Usually, one intuitively motivates the double integral by cutting the plane by little ...


0

Try this. In 3 dimensions, the cross product of (a1, a2, a3) and (b1, b2, b3) is (a2*b3- a3*b2, a3*b1- a1*b3, a1*b2- a2*b1) Now try the inner product of a and this vector, the inner product of b and this vector. This cross product is orthogonal (perpendicular) to each of these. Since two vectors in 3 dimensions define a plane (unless the two are ...


-1

The cross-product was defined by Gibbs to mimic the properties of quaternion multiplication without the negativity of the square. Quaternions are very economic way to represent rotations in 3D and therefore many physical equations. See for example here. http://en.wikipedia.org/wiki/Geometric_algebra http://en.wikipedia.org/wiki/History_of_quaternions ...


3

Its not really an "intuition" thing, the cross product is defined that way. The following may help you understand why. The short answer is that the cross product appears in many physical systems - most famously in calculating the force produced by an electrical current in a magnetic field. But this begs the question of why it appears so often in physics. ...



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