Question

Value of the scalar triple product \(\vec{a}\cdot(\vec{b}\times\vec{c})\) is (answer in integer) ______.

\[ \vec{a}=2\hat{i}-3\hat{j}+4\hat{k} \]
\[ \vec{b}=\hat{i}+2\hat{j}-3\hat{k} \]
\[ \vec{c}=3\hat{i}+4\hat{j}-\hat{k} \]
Here, \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) are mutually orthogonal unit vectors.

Hint:

First find b cross c using the determinant method, then take its dot product with a.

Answer 1

Correct Answer: 36

Step 1: Use the direct determinant shortcut.
Rather than computing $\vec{b}\times\vec{c}$ first and then dotting with $\vec{a}$, the scalar triple product $\vec{a}\cdot(\vec{b}\times\vec{c})$ equals a single 3 by 3 determinant with the three vectors placed as rows in order.

Step 2: Set up the determinant.
\[ \vec{a}\cdot(\vec{b}\times\vec{c})=\begin{vmatrix}2&-3&4\\1&2&-3\\3&4&-1\end{vmatrix} \]

Step 3: Expand along the first row.
\[ =2\begin{vmatrix}2&-3\\4&-1\end{vmatrix}-(-3)\begin{vmatrix}1&-3\\3&-1\end{vmatrix}+4\begin{vmatrix}1&2\\3&4\end{vmatrix} \]

Step 4: Work out each 2 by 2 minor.
$\begin{vmatrix}2&-3\\4&-1\end{vmatrix}=(2)(-1)-(-3)(4)=-2+12=10$
$\begin{vmatrix}1&-3\\3&-1\end{vmatrix}=(1)(-1)-(-3)(3)=-1+9=8$
$\begin{vmatrix}1&2\\3&4\end{vmatrix}=(1)(4)-(2)(3)=4-6=-2$

Step 5: Combine the terms.
\[ =2(10)+3(8)+4(-2)=20+24-8=36 \]

Step 6: Conclude.
Both routes must agree, since the triple product is a single fixed number for given vectors.
\[ \boxed{36} \]