12.5.4 Find parametric equations .

[karush]

$\textsf{Find parametric equations .}$
$\textsf{of the line through the point }$
$$P(-3, -4, -2)$$
$\textsf{and perpendicular to the vectors }$
$$u = -6i + 2j + 8k$$and $$v = -7i + 5 j - 2k$$
$\textit{Answer:$\displaystyle x = -44t - 3 , y = -68t - 4, z = -16t - 2 $} $
ok how is this done with 2 vectors

 

[Evgeny.Makarov]

You can find a vector perpendicular to $(-6,2,8)$ and $(-7,5,-2)$ by computing the vector product of these vectors or by finding any nontrivial solution to the system of equations \(\displaystyle \left\{\begin{align}-6x+2y+8z&=0\\-7x+5y-2z&=0\end{align}\right.\).

 

[karush]

$\displaystyle \begin{bmatrix}
i & j & k\\ -6 &2 &8\\ -7 &5 &-2 \end{bmatrix}
= \begin{bmatrix} 2&8\\ 5&-2 \end{bmatrix}i
- \begin{bmatrix}-6&8\\-7&-2 \end{bmatrix}j
+ \begin{bmatrix}-6&2\\-7&5 \end{bmatrix}k
=-44i-68j-16k\\$
$\text{so the parametric would be from P(-3,-4,-2)}$
\begin{align*}
x&=-44t-3,y=-68t-4,z=-16t-2
\end{align*}

 

[Evgeny.Makarov]

The sign before the determinant multiplied by $j$ should be a minus. The equation for $x$ should be $x=-44t-3$ and for $z$ it should be $z=-16t-2$. You can multiply all coordinates of the vector (cross) product by any nonzero number, for example, by $-1/4$.

 

[karush]

The sign before the determinant multiplied by $j$ should be a minus. The equation for $x$ should be $x=-44t-3$ and for $z$ it should be $z=-16t-2$. You can multiply all coordinates of the vector (cross) product by any nonzero number, for example, by $-1/4$.

Well one thing nice here at MHB is the typos are shown
other forums rarely do that

however the book answers have had wrong answers at times