+0  
 
0
71
1
avatar

a) Let $\mathbf{p} =  \begin{pmatrix} 2\\ -1\\ 2 \end{pmatrix}$ and $\mathbf{d} =\begin{pmatrix} 1\\ -1\\ 4 \end{pmatrix}$ and consider the particle that visits the point $(x,y,z)$ at time $t$ if

\[\begin{pmatrix} x \\ y \\ z \end{pmatrix} =  \mathbf{p} + t\mathbf{d}.\]The vectors $\mathbf{p}$ and $\mathbf{d}$ and the graph of the parametrization are shown below:


If the intersection between this line and the $xy$-plane is $(a,b,c),$ enter $a, b, c$ in this order below.

 

b) 

Let $\mathbf{p} = \begin{pmatrix}-1\\ -2 \\ 1\end{pmatrix}$ and $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and consider the particle that visits $(x,y,z)$ at time $t$ if
\[\begin{pmatrix}x \\ y \\ z \end{pmatrix} = \mathbf{p} + t\mathbf{d}.\]The vectors $\mathbf{p}$ and $\mathbf{d}$ and the graph of the parametrization are shown below:

 


Then if $R = (-1, 3, 5)$, calculate the point $(a,b,c)$ on the line that is closest to $R$. Enter $a, b, c$ in that order below.

 

help asap

 Feb 10, 2021
edited by Guest  Feb 10, 2021
 #1
avatar
0

(a) The intersection is (2,3,-2).

 

(b) The answer is (3,-1,5).

 Feb 10, 2021

26 Online Users

avatar
avatar