Vector Algebra
Parallelepiped Volume
Grade 12
Question:
<p>The value of a so that the volume of the parallelepiped formed by <span style='font-weight:bold;'>i</span> + a<span style='font-weight:bold;'>j</span> + <span style='font-weight:bold;'>k</span>, <span style='font-weight:bold;'>j</span> + a<span style='font-weight:bold;'>k</span>, and a<span style='font-weight:bold;'>i</span> + <span style='font-weight:bold;'>k</span> becomes minimum is</p>
<p>(a) -3</p>
<p>(b) 3</p>
<p>(c) 1/√3</p>
<p>(d) √3</p>
Step-by-Step Solution
Key Concept: The volume of a parallelepiped is given by the absolute value of the scalar triple product. To minimize, express the volume as a function of a and use calculus to find the critical point.
Volume of the parallelepiped \(V = (\mathbf{i} + a\mathbf{j} + \mathbf{k}) \cdot [(\mathbf{j} + a\mathbf{k}) \times (a\mathbf{i} + \mathbf{k})]\) This is the scalar triple product of the three vectors. To find the minimum, we take the derivative with respect to a and set it equal to zero. The minimum occurs at \(a = \frac{1}{\sqrt{3}}\). ∴ Answer is (c) 1/√3.
Correct Answer: C