Question

problem 6
(fill in the blank) : calculate the shortest distance between the point,
$q = (1, 3, 1)$,
and the plane given by the vector equation,
$\vec{r} = \vec{p} + s\vec{v} + t\vec{w}$,
where
$\vec{p} = \langle 1, 0, 1 \
angle$
and
$\vec{v} = \langle -2, 1, 3 \
angle$
and
$\vec{w} = \langle 1, 2, 1 \
angle$
and
$s, t \in \mathbb{r}$.
for full credit, please show the relevant calculations.
distance = <your answer here>

Explanation:

Step1: Find plane normal vector

Compute cross product $\vec{v} \times \vec{w}$:

$$ \vec{v} \times \vec{w} =
LATEXBLOCK0
= \hat{i}(1\cdot1 - 3\cdot2) - \hat{j}(-2\cdot1 - 3\cdot1) + \hat{k}(-2\cdot2 - 1\cdot1) = \langle -5, 5, -5 angle $$

Step2: Get plane scalar equation

Use point $\vec{p}=(1,0,1)$ and normal $\langle -5,5,-5
angle$:

$$ -5(x-1) + 5(y-0) -5(z-1) = 0 $$

Simplify to: $x - y + z - 2 = 0$ (divide by -5)

Step3: Define point and plane values

For $Q=(1,3,1)$ and plane $ax+by+cz+d=0$ (here $a=1, b=-1, c=1, d=-2$):

$$ \vec{PQ} = \langle 1-1, 3-0, 1-1 angle = \langle 0,3,0 angle $$

Step4: Compute shortest distance

Use distance formula $\frac{|aQ_x + bQ_y + cQ_z + d|}{\sqrt{a^2+b^2+c^2}}$:

$$ \text{Distance} = \frac{|1\cdot1 + (-1)\cdot3 + 1\cdot1 - 2|}{\sqrt{1^2+(-1)^2+1^2}} = \frac{|1-3+1-2|}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} $$

Answer:

$\sqrt{3}$