From the given equation:
\[ \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}. \]
Expanding using vector algebra:
\[ \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}. \]
It is given:
\[ \vec{a} \times \vec{b} = \hat{i} + 8\hat{j} + 13\hat{k}. \]
So:
\[ \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \vec{0}. \]
Expanding further:
\[ \vec{b} \times \vec{c} = -\vec{a} \times \vec{c}. \]
Using \( \vec{a} \cdot \vec{c} = 13 \), compute:
\[ \vec{b} \cdot \vec{c} = -\left[\vec{a} \cdot (\hat{i} + 8\hat{j} + 13\hat{k})\right] = -22. \]
From the determinant of \( \vec{b} \cdot \vec{c} \):
\[ 24 - \vec{b} \cdot \vec{c} = 46. \]
Let \( \vec{a} \) and \( \vec{b} \) be two co-initial vectors forming adjacent sides of a parallelogram such that:
\[
|\vec{a}| = 10, \quad |\vec{b}| = 2, \quad \vec{a} \cdot \vec{b} = 12
\]
Find the area of the parallelogram.
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).