Q66 / 180 · COMEDK 2021 · Mathematics
If for $\mathrm{a}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \mathrm{b}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\mathrm{c}=-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$, then find [a b c $]$.
- A$-15$
- B$-10$
- Correct answerC$-30$
- D$-5$
Solution
The scalar triple product $[\mathrm{a} \mathrm{b} \mathrm{c}]$ is given by the determinant of the matrix formed by the components of the vectors $\mathrm{a}$, $\mathrm{b}$, and $\mathrm{c}$.
$[\mathrm{a} \mathrm{b} \mathrm{c}] = \begin{vmatrix} 2 & 3 & 1 \\ 1 & -2 & 1 \\ -3 & 1 & 2 \end{vmatrix}$
Expanding the determinant along the first row:
$[\mathrm{a} \mathrm{b} \mathrm{c}] = 2((-2)(2) - (1)(1)) - 3((1)(2) - (1)(-3)) + 1((1)(1) - (-2)(-3))$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = 2(-4 - 1) - 3(2 + 3) + 1(1 - 6)$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = 2(-5) - 3(5) + 1(-5)$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = -10 - 15 - 5$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = -30$
Answer: $-30$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = \begin{vmatrix} 2 & 3 & 1 \\ 1 & -2 & 1 \\ -3 & 1 & 2 \end{vmatrix}$
Expanding the determinant along the first row:
$[\mathrm{a} \mathrm{b} \mathrm{c}] = 2((-2)(2) - (1)(1)) - 3((1)(2) - (1)(-3)) + 1((1)(1) - (-2)(-3))$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = 2(-4 - 1) - 3(2 + 3) + 1(1 - 6)$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = 2(-5) - 3(5) + 1(-5)$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = -10 - 15 - 5$
$[\mathrm{a} \mathrm{b} \mathrm{c}] = -30$
Answer: $-30$
Hence, the correct answer is option: 'C'.