Αποτελέσματα Αναζήτησης
14 Οκτ 2017 · Unfortunately, your two chosen vectors are parallel - all three points are on a line in the plane - so their cross-product will necessarily be zero; if you choose one of your three points to not lie in the same line in the plane then the cross-product will be parallel to your extracted normal vector.
11 Μαρ 2018 · Plugging the coordinates of the known points into this generic equation gives you the following system of linear equations: a − 2b + c + d = 0 4a − 2b − 2c + d = 0 4a + b + 4c + d = 0. Solve this system for the unknown coefficients a, b, c and d. The solution won’t be unique, but if all goes well (you haven’t made a mistake and the ...
Okay, so we are given the two direction vectors. The equation of a plane is stated based on its normal. so we take the cross product of the two direction vectors to get a vector perpendicular to both.
5 Μαΐ 2016 · However, the solution gives the vector equation as: $(x,y,z)=(1,0,0)+\lambda(3,1,0)+t(4,0,1)$. I understand that there are multiple ways to find the vector equation of a plane. But since I am doing this for transformation purposes, the vector equation I found is a little more complicated than the solution's equation.
18 Μαρ 2015 · The vector equation of a plane has the form $\mathbf r(s,t) = \mathbf r_0 + s\mathbf u + t\mathbf v$. To find $\mathbf u$ and $\mathbf v$, you need two vectors (which are not collinear) which are orthogonal to $(1,2,7)$ (do you see why they need to be orthogonal to this vector?).
8 Ιαν 2016 · I would like to visualize planes in 3D as I start learning linear algebra, to build a solid foundation. Surprisingly, I have been unable to find an online tool (website/web app) to visualize planes in 3 dimensions. For example, I'd like to be able to enter 3 points and see the plane. Does something like this exist?
This one is really giving me a hard time. I know that to find the plane perpendicular to the line I can use the vector n between two points on the line and and the plane. I cannot wrap my mind around how to reverse this process, particularly because the plane is equal to 1 and not zero. Any help would be greatly appreciated.
28 Οκτ 2019 · Note that the intersection point has to satisfy both conditions, so it is enouh to plug in the line form into the plane equation and solve: $$(P_v+\lambda \vec{v_v}-P_p)\cdot\vec{v_p}=0 \iff \lambda =\frac{(P_p-P_v)\cdot\vec{v_p}}{\vec{v_v}\cdot\vec{v_p}}$$ Of course, if $\vec{v_v}\cdot\vec{v_p}=0$, both elements would be parallel, so there would not be any intersection point.
6 Ιουν 2018 · Let r r → be the position vector of any point in the plane. let p p → be the position vector of the point of intersection of the two (non parallel) lines that have been given. Clearly r −p r → − p → lies in the plane, hence it is perpendicular to the normal to the plane (given by the cross product of 2i − j + 3k 2 i − j + 3 k ...
13 Ιουν 2015 · If $\mathbf{n}$ is the normal vector to the given plane and $\mathbf{p}$ is the point through which the line is supposed to pass, then the equation of the line will be of the form $\mathbf{r}=\mathbf{p}+t\mathbf{n}$.
