Recent content by GPPaille

1- Take an easier example, like the rectangular inclined parallelogram define by the two vectors \vec{A}=(a,0,0), \vec{B}=(0,b,c). Only two projections has non-null area. Defining h=\sqrt{b^2 + c^2}, we know that P = ah since it's rectangular. But P = ah = a\sqrt{b^2 + c^2} = \sqrt{(ab)^2...

You need to define (x,y,z) with (r,t,z'): x = \frac{r}{2}cos\theta y = \frac{r}{3}sin\theta z = z' \left| J \right| = \begin{vmatrix} \frac{1}{2}cos\theta & -\frac{r}{2}sin\theta & 0 \\ \frac{1}{3}sin\theta & \frac{r}{3}cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} =...

Sorry, I don't have any picture, but it's simple to draw it on paper. Just take the YZ-plane projection and draw the two vectors (a_y,a_z), (b_y,b_z). It's like when you look at it from the X axis. \overrightarrow{i} \times \overrightarrow{k} = - \overrightarrow{j} because it's the algebraic...

(a)^{\frac{1}{x}}=(b^x)^{\frac{1}{x}} a^{\frac{1}{x}}=b

Yes you can do it. It's even simpler than what I said but it's exactly the same thing, except that it's in one single step. So you have your answer or your stuck there?

0 = 0 + 0 = (2x1 + y1 + 3z1) + (2x2 + y2 + 3z2) = 2(x1+x2) + (y1 + y2) + 3(z1 + z2) If you write x3=x1+x2, y3=y1+y2, z3=z1+z2, then it will become clear that this vector (x3,y3,z3) is also in W.

The answer is correct but the solution is guesswork. The goal of this exercise is to be able to create a mathematical model for the problem. You have three kinds of day. 1- Walk, 2- Badminton, 3- Nothing. These are your three unknowns, there's also three equations in the question. It will be...

You should try to create a system of linear equations. Define a variable for each "activity", and do the sum. If you do nothing in a morning, it doesn't say that you did nothing in the evening. Your answer is wrong.

ok! So we need to do it in three steps: 1- Show that the area of a parallelogram in 2D is the determinant of a 2x2 matrix containing the two side vectors. 2- Show that the area of a parallelogram in 3D is the squared sum of his projection on the three axis planes. 3- Make the final link...

Then I really don't see what you want to know. I proved by an algebraic property that |c|=|a||b|sin\theta and I did not suppose anything.

Using Lagrange's identity identity, you get |a \times b|^2 = (a \cdot a)(b \cdot b) - (a \cdot b)(a \cdot b) Which is simply equal to |a \times b|^2 = |a|^2|b|^2 - |a|^2|b|^2cos^2(\theta) |a \times b|^2 = |a|^2|b|^2 (1 - cos^2(\theta)) |a \times b|^2 = |a|^2|b|^2 sin^2(\theta)

But you don't have to apply T twice to see that (1,0) do not preserves length. Write it in term of the basis vectors. And you will see that only one basis vector preserves its length.

\left\|(\sqrt{2},i)\right\| = (\sqrt{2},i) \cdot (\sqrt{2},i)^* = 3

Here, p is a scalar, that comes magically from the projection. And if it don't use the dot product to compute p, then I don't see the link between the dot product and cos(theta).

But the dot product is hidden in |p|, and the way we compute the projection using the dot product uses the fact that we know that it's equal to cos(theta)/|A||B|. Explicitly, it looks like this: 1- Hypothesis: A dot B = cos(theta)/|A||B| 2- Use it to define projection 3- Use projection...