Vector Algebra
Scalar Triple Product
Grade None
Question:
<p>If <span style='font-weight:bold;'>a</span> = <span style='font-weight:bold;'>i</span> + <span style='font-weight:bold;'>j</span> + <span style='font-weight:bold;'>k</span>, <span style='font-weight:bold;'>b</span> = <span style='font-weight:bold;'>i</span> - <span style='font-weight:bold;'>j</span> + <span style='font-weight:bold;'>k</span>, <span style='font-weight:bold;'>c</span> = <span style='font-weight:bold;'>i</span> + 2<span style='font-weight:bold;'>j</span> - <span style='font-weight:bold;'>k</span>, then the value of \[\begin{vmatrix} \mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c} \end{vmatrix}\] is</p>
<p>(a) 2</p>
<p>(b) 4</p>
<p>(c) 16</p>
<p>(d) 64</p>
Step-by-Step Solution
Key Concept: The given determinant of dot products equals the square of the scalar triple product [a b c]². Use the formula to compute the scalar triple product from the component form of vectors.
We have, a = i + j + k , b = i - j + k , c = i + 2 j - k We know that, \[\begin{vmatrix} \mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c} \end{vmatrix} = [\mathbf{a} \ \mathbf{b} \ \mathbf{c}]^2\] \[= \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix}^2\] \[= [1(1 - 2) - 1(-1 - 1) + 1(2 + 1)]^2\] \[= [-1 + 2 + 3]^2 = [4]^2 = 16\] ∴ Answer is (c) 16.
Correct Answer: C