To find the area of the parallelogram with one vertex at [tex]\( P \)[/tex] and sides formed by the vectors [tex]\( \overrightarrow{PQ} \)[/tex] and [tex]\( \overrightarrow{PR} \)[/tex], we can follow these steps:
1. Find the vectors [tex]\( \overrightarrow{PQ} \)[/tex] and [tex]\( \overrightarrow{PR} \)[/tex]:
[tex]\[
P = (0, 1, 1), \quad Q = (2, 0, -4), \quad R = (-3, -2, 1)
\][/tex]
[tex]\[
\overrightarrow{PQ} = Q - P \implies (2 - 0, 0 - 1, -4 - 1) = (2, -1, -5)
\][/tex]
[tex]\[
\overrightarrow{PR} = R - P \implies (-3 - 0, -2 - 1, 1 - 1) = (-3, -3, 0)
\][/tex]
2. Calculate the cross product [tex]\( \overrightarrow{PQ} \times \overrightarrow{PR} \)[/tex]:
The cross product can be found using the determinant of a matrix composed of the unit vectors and the components of [tex]\( \overrightarrow{PQ} \)[/tex] and [tex]\( \overrightarrow{PR} \)[/tex]:
[tex]\[
\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -1 & -5 \\
-3 & -3 & 0
\end{vmatrix}
\][/tex]
Expanding this determinant, we get:
[tex]\[
\overrightarrow{PQ} \times \overrightarrow{PR} = \mathbf{i}((-1)(0) - (-5)(-3)) - \mathbf{j}((2)(0) - (-5)(-3)) + \mathbf{k}((2)(-3) - (-1)(-3))
\][/tex]
[tex]\[
= \mathbf{i}(0 - 15) - \mathbf{j}(0 - 15) + \mathbf{k}(-6 - 3)
\][/tex]
[tex]\[
= \mathbf{i}(-15) - \mathbf{j}(-15) + \mathbf{k}(-9)
\][/tex]
[tex]\[
= -15\mathbf{i} + 15\mathbf{j} - 9\mathbf{k}
\][/tex]
Thus, the cross product is:
[tex]\[
\overrightarrow{PQ} \times \overrightarrow{PR} = (-15, 15, -9)
\][/tex]
3. Find the magnitude of the cross product [tex]\( \lVert \overrightarrow{PQ} \times \overrightarrow{PR} \rVert \)[/tex]:
The magnitude of a vector [tex]\((a, b, c)\)[/tex] is given by:
[tex]\[
\lVert (a, b, c) \rVert = \sqrt{a^2 + b^2 + c^2}
\][/tex]
Applying this to our cross product vector:
[tex]\[
\lVert (-15, 15, -9) \rVert = \sqrt{(-15)^2 + 15^2 + (-9)^2}
\][/tex]
[tex]\[
= \sqrt{225 + 225 + 81}
\][/tex]
[tex]\[
= \sqrt{531}
\][/tex]
4. Understand that the area of the parallelogram is given by the magnitude of the cross product:
Thus, the area of the parallelogram is:
[tex]\[
\sqrt{531}
\][/tex]
After simplification, this becomes:
[tex]\[
\text{The area of the parallelogram is } 3\sqrt{59}.
\][/tex]
