Question

problem 5
(fill in the blank) : write the scalar equation for a plane containing the point,
$q = (2, 4, 3)$,
and the line given by the vector equation,
$\vec{r_l} = \vec{p} + s\vec{v}$,
where
$\vec{p} = \hat{i} + 2\hat{j} + 4\hat{k}$
and
$\vec{v} = 2\hat{i} + 3\hat{j} + 2\hat{k}$
and
$s \in \mathbb{r}$.
for full credit, please show the relevant calculations.
<your answer here>

Explanation:

Step1: Get point on line

From $\vec{p} = \hat{i} + 2\hat{j} + 4\hat{k}$, point $P=(1,2,4)$.

Step2: Find vector $\overrightarrow{PQ}$

$\overrightarrow{PQ} = Q - P = (2-1, 4-2, 3-4) = (1,2,-1)$

Step3: Identify direction vector $\vec{v}$

$\vec{v} = (2,3,2)$

Step4: Compute normal vector $\vec{n}$

$\vec{n} = \overrightarrow{PQ} \times \vec{v} =

$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 3 & 2 \end{vmatrix}$$

= \hat{i}(4+3) - \hat{j}(2+2) + \hat{k}(3-4) = 7\hat{i} -4\hat{j} -1\hat{k}$

Step5: Write plane equation

Use point $Q=(2,4,3)$: $7(x-2) -4(y-4) -1(z-3)=0$
Simplify: $7x -14 -4y +16 -z +3=0$
$7x -4y -z +5=0$

Answer:

$7x - 4y - z + 5 = 0$