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# Let $\mathbf{p} = \begin{pmatrix} 2\\ -1\\ 2 \end{pmatrix}$ and $\mathbf{d} =\begin{pmatrix} 1\\ -1\\ 4 \end{pmatrix}$ and consider the par

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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