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Magazine Chapter 1: Problem 38
Consider the parallelepiped with edges \(O A, O B\), and \(O C\), where \(A(2,1,0), B(1,2,0)\), and \(C(0,1, \alpha)\). a. Find the real number \(\alpha>0\) such that the volume of the parallelepiped is 3 units \(^{3}\). b. For \(\alpha=1\), find the height \(h\) from vertex \(C\) of the parallelepiped. Sketch the parallelepiped.
Short Answer
Expert verified
a) \( \alpha = 1 \). b) Height \( h = 1 \) unit.
Step by step solution
01
Understand the Parallelepiped Volume Formula
The volume of a parallelepiped with edges defined by vectors \( \vec{OA}, \vec{OB} \), and \( \vec{OC} \) can be calculated using the scalar triple product: \( V = |\vec{OA} \cdot ( \vec{OB} \times \vec{OC} )| \). Each vector is the position vector of the given points \( A(2,1,0), B(1,2,0), \) and \( C(0,1, \alpha) \) relative to the origin \( O(0,0,0) \). Write the vectors as follows: \( \vec{OA} = (2, 1, 0) \), \( \vec{OB} = (1, 2, 0) \), and \( \vec{OC} = (0, 1, \alpha) \).
02
Calculate the Cross Product \( \vec{OB} \times \vec{OC} \)
Calculate \( \vec{OB} \times \vec{OC} \) using the determinant method:\[\vec{OB} \times \vec{OC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 0 \ 0 & 1 & \alpha \end{vmatrix} = \mathbf{i}(2\alpha) - \mathbf{j}(\alpha) + \mathbf{k}(1) = (2\alpha, -\alpha, 1).\]
03
Compute the Scalar Triple Product
Now find the dot product \( \vec{OA} \cdot ( \vec{OB} \times \vec{OC} ) \):\[\vec{OA} \cdot ( \vec{OB} \times \vec{OC} ) = (2, 1, 0) \cdot (2\alpha, -\alpha, 1) = 2(2\alpha) + 1(-\alpha) + 0(1) = 4\alpha - \alpha = 3\alpha.\]
04
Solve for \( \alpha \) Given the Volume
Since the volume is given as 3 units\(^3\), set the equation equal to 3:\[ |3\alpha| = 3. \]Since \(\alpha > 0\), this simplifies to:\[ 3\alpha = 3 \quad \implies \quad \alpha = 1. \]
05
Determine the Height \( h \) from Vertex \( C \)
Given \( \alpha = 1 \), the height \( h \) from vertex \( C(0,1,1) \) is measured perpendicularly to the plane formed by vectors \( \vec{OA} \) and \( \vec{OB} \). The normal vector to this plane is \( \vec{n} = \vec{OA} \times \vec{OB} \). Compute this:\[\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 1 & 0 \ 1 & 2 & 0 \end{vmatrix} = \mathbf{k}(4 - 1) = (0, 0, 3).\]The height \( h \) is the absolute value of the projection of \( \vec{OC} \) onto \( \vec{n} \):\[h = \frac{|\vec{OC} \cdot \vec{n}|}{||\vec{n}||} = \frac{|(0,1,1) \cdot (0,0,3)|}{3} = \frac{3}{3} = 1.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Scalar Triple Product
The Scalar Triple Product is a useful operation to find the volume of a parallelepiped. It involves three vectors and results in a scalar, which is the volume's numerical value. For vectors \(\vec{a}, \vec{b},\) and \(\vec{c}\), the scalar triple product \(\vec{a} \cdot (\vec{b} \times \vec{c})\) gives the volume of the parallelepiped formed by these vectors when placed tail to head. Here鈥檚 how this works:
- The operation starts with calculating the cross product \(\vec{b} \times \vec{c}\), resulting in a vector perpendicular to both \(\vec{b}\) and \(\vec{c}\).
- Next, the dot product \(\vec{a} \cdot (\vec{b} \times \vec{c})\) involves projecting \(\vec{a}\) onto the resultant vector from the cross product.
- The absolute value \(|\vec{a} \cdot (\vec{b} \times \vec{c})|\) represents the volume of the parallelepiped.
Cross Product
The cross product is a fundamental operation in vector algebra involving two vectors. It results in a third vector that is perpendicular to the plane of the initial vectors. This property makes the cross product ideal for problems involving perpendicularity and volume.To perform a cross product for vectors \(\vec{u} = (a, b, c)\) and \(\vec{v} = (d, e, f)\), the calculation involves the following determinant:\[\vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a & b & c \ d & e & f \end{vmatrix} = \mathbf{i}(bf - ce) - \mathbf{j}(af - cd) + \mathbf{k}(ae - bd)\]This results in a new vector \((bf - ce, af - cd, ae - bd)\). In our exercise, the specific calculation was \(\vec{OB} \times \vec{OC}\), which determined the perpendicular vector crucial for finding the scalar triple product.The cross product is peculiar in that it is only defined in three dimensions, being inseparably linked to the geometric interpretation of vectors in three-dimensional space.
Projection of a Vector
The projection of a vector is a crucial concept in understanding geometrical relationships between vectors, especially when measuring heights or distances perpendicular to a plane. In essence, projecting a vector \(\vec{a}\) onto another vector \(\vec{b}\) finds how much of \(\vec{a}\) goes in the direction of \(\vec{b}\).For two vectors \(\vec{u}\) and \(\vec{v}\), the scalar projection is given by\[\mathrm{Proj}_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{||\vec{v}||}\]The vector projection is then\[\left(\frac{\vec{u} \cdot \vec{v}}{||\vec{v}||^2}\right) \vec{v}\]When you're finding the height in our exercise, you project \(\vec{OC}\) onto the normal vector \(\vec{n}\). The formula gives how far \(\vec{OC}\) extends in the direction perpendicular to the base plane formed by \(\vec{OA}\) and \(\vec{OB}\). This projection simplifies determining the height \(h\) from the vertex of the parallelepiped using the constructed normal vector, resulting in the equation \(h = \frac{|\vec{OC} \cdot \vec{n}|}{||\vec{n}||}\). By calculating this, you find the perpendicular distance crucial for the right measurement of space.
