Question

problem 2
(fill in the blank): to answer parts 1 and 2, use the following vector equation for a plane,
$vec{r} = vec{p} + svec{v} + tvec{w}$,
where
$vec{r} = langle x, y, z \
angle$
and
$vec{p} = langle 1, 3, 1 \
angle$
and
$vec{v} = langle -2, -3, 2 \
angle$
and
$vec{w} = langle 3, 3, -4 \
angle$
for all
$s, t in mathbb{r}$.

problem 2 - part 1
(fill in the blank): write the scalar equation for the plane.
for full credit, please show the relevant calculations.
<your answer here>

problem 2 - part 2

Explanation:

Step1: Find normal vector via cross product

Compute $\vec{n} = \vec{v} \times \vec{w}$

$$ \vec{n} =
LATEXBLOCK0
= \boldsymbol{i}[(-3)(-4)-(2)(3)] - \boldsymbol{j}[(-2)(-4)-(2)(3)] + \boldsymbol{k}[(-2)(3)-(-3)(3)] = \boldsymbol{i}(12-6) - \boldsymbol{j}(8-6) + \boldsymbol{k}(-6+9) = \langle 6, -2, 3 angle $$

Step2: Use point $\vec{p}$ for scalar equation

Apply $\vec{n} \cdot (\vec{r} - \vec{p}) = 0$

$$ 6(x-1) - 2(y-3) + 3(z-1) = 0 $$

Step3: Expand and simplify the equation

$$ 6x - 6 - 2y + 6 + 3z - 3 = 0 \\ 6x - 2y + 3z - 3 = 0 \\ 6x - 2y + 3z = 3 $$

Answer:

$6x - 2y + 3z = 3$