Question

let ( p = (0,1,0) ), ( q = (1,1,1) ), ( r = (-1,-1,-1) ). find (a) the area of the triangle ( pqr ). area = (b) the equation for a plane that contains ( p ), ( q ), and ( r ). this plane is submit answer next item

Explanation:

Part (a)

Step1: Find vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$

$\overrightarrow{PQ} = Q - P = (1 - 0, 1 - 1, 1 - 0) = (1, 0, 1)$
$\overrightarrow{PR} = R - P = (-1 - 0, -1 - 1, -1 - 0) = (-1, -2, -1)$

Step2: Compute the cross product $\overrightarrow{PQ} \times \overrightarrow{PR}$

$$ \overrightarrow{PQ} \times \overrightarrow{PR} =
LATEXBLOCK0
= \mathbf{i}(0 \cdot (-1) - 1 \cdot (-2)) - \mathbf{j}(1 \cdot (-1) - 1 \cdot (-1)) + \mathbf{k}(1 \cdot (-2) - 0 \cdot (-1)) = \mathbf{i}(2) - \mathbf{j}(0) + \mathbf{k}(-2) = (2, 0, -2) $$

Step3: Find the magnitude of the cross product

$|\overrightarrow{PQ} \times \overrightarrow{PR}| = \sqrt{2^2 + 0^2 + (-2)^2} = \sqrt{4 + 0 + 4} = \sqrt{8} = 2\sqrt{2}$

Step4: Calculate the area of the triangle

The area of $\triangle PQR$ is $\frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}| = \frac{1}{2} \times 2\sqrt{2} = \sqrt{2}$

Part (b)

Step1: Use the normal vector from the cross product

From part (a), the normal vector $\mathbf{n}$ to the plane is $(2, 0, -2)$ (or we can simplify it to $(1, 0, -1)$ by dividing by 2).

Step2: Use point $P(0, 1, 0)$ to find the plane equation

The general plane equation is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$, where $(x_0, y_0, z_0)$ is a point on the plane and $(a, b, c)$ is the normal vector.
Using $P(0, 1, 0)$ and $\mathbf{n} = (1, 0, -1)$:
$1(x - 0) + 0(y - 1) - 1(z - 0) = 0$
Simplify: $x - z = 0$ or $x = z$

Answer:

s:
(a) $\sqrt{2}$
(b) $x - z = 0$ (or equivalent forms like $x = z$)