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: Recall the meaning of the scalar triple product.
For three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), the scalar triple product \(\vec{a}\cdot(\vec{b}\times\vec{c})\) is found by first finding the cross product \(\vec{b}\times\vec{c}\), which gives a new vector, and then taking its dot product with \(\vec{a}\).

Step 2: Write down the given vectors.
\[ \vec{a}=2\hat{i}-3\hat{j}+4\hat{k} \]
\[ \vec{b}=\hat{i}+2\hat{j}-3\hat{k} \]
\[ \vec{c}=3\hat{i}+4\hat{j}-\hat{k} \]

Step 3: Find \(\vec{b}\times\vec{c}\) using the determinant form.
\[ \vec{b}\times\vec{c}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&-3\\3&4&-1\end{vmatrix} \]
Expand along the first row:
\[ \hat{i}\big[(2)(-1)-(-3)(4)\big]-\hat{j}\big[(1)(-1)-(-3)(3)\big]+\hat{k}\big[(1)(4)-(2)(3)\big] \]
\[ =\hat{i}(-2+12)-\hat{j}(-1+9)+\hat{k}(4-6) \]
\[ =10\hat{i}-8\hat{j}-2\hat{k} \]

Step 4: Take the dot product with \(\vec{a}\).
\[ \vec{a}\cdot(\vec{b}\times\vec{c})=(2)(10)+(-3)(-8)+(4)(-2) \]
\[ =20+24-8 \]
\[ =36 \]

Step 5: Final answer.
\[ \boxed{36} \]