The problem given was "Given a the vector equation $r(t)=(−4−1t)i+(−3+1t)j+(3+5t)k$, rewrite this in terms of the symmetric equations for the line."
(quotient involving $x$)$=\quad -x- 4$
(quotient involving $y$)$=\quad\frac{y+3}{1}$
(quotient involving $z$) ?
I am having trouble figuring out the quotient involving $z$. I intially got $\frac{z-3}{5}$ but that was wrong. I googled how to do problems similiar to this one and it said to write them as parametric equations and then solve for $t$ which worked for the other two. I don't know what I am doing wrong and any help would be appreciated. Thank you !
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$\begingroup$Re-write $\vec r=(-4\vec i-3\vec j+3\vec k)+t(-1\vec i+\vec j+ 5\vec k)$So this line passes through the point $(-4,-3,3)$ having direction ratios as $(-1,1,5).$ The equation of liner is:$$\frac{x+4}{-1}=\frac{y+3}{1}=\frac{z-3}{5}=t.$$
$\endgroup$ $\begingroup$In 3-dimensional space, a line's vector equation has the following form, right?$$\begin{pmatrix} x \cr y \cr z \end{pmatrix}=\begin{pmatrix} a_x \cr a_y \cr a_z \end{pmatrix}+t\begin{pmatrix} u_x \cr u_y \cr u_z \end{pmatrix}$$
Rearranging:$$\begin{pmatrix} x-a_x \cr y-a_y \cr z-a_z \end{pmatrix}=t\begin{pmatrix} u_x \cr u_y \cr u_z \end{pmatrix}.$$
Dividing and multiplying:$$\begin{pmatrix} \frac{x-a_x}{u_x} \cr \frac{y-a_y}{u_y} \cr \frac{z-a_z}{u_z} \end{pmatrix}=\begin{pmatrix} t \cr t \cr t \end{pmatrix}.$$
Eliminating $t$ to obtain the symmetric form:$$\frac{x-a_x}{u_x} = \frac{y-a_y}{u_y} = \frac{z-a_z}{u_z}.$$
Therefore, in your example,$$\mathbf{r}(t)=(−4−1t)\mathbf{i}+(−3+1t)\mathbf{j}+(3+5t)\mathbf{k} \\ \implies \frac{x+4}{-1} = \frac{y+3}{1} = \frac{z-3}{5}.$$
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