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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]
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]
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