Step 1: Given,
\( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} \Rightarrow \vec{a} \cdot (\vec{b} - \vec{c}) = 0 \)
So, \( \vec{a} \) is perpendicular to \( (\vec{b} - \vec{c}) \).
Step 2: Given,
\( \vec{a} \times \vec{b} = \vec{a} \times \vec{c} \Rightarrow \vec{a} \times (\vec{b} - \vec{c}) = \vec{0} \)
So, \( \vec{a} \) is parallel to \( (\vec{b} - \vec{c}) \).
Step 3:
So we have:
- \( \vec{b} - \vec{c} \) is perpendicular to \( \vec{a} \)
- \( \vec{b} - \vec{c} \) is also parallel to \( \vec{a} \)
The only vector that can be both perpendicular and parallel to the same non-zero vector is the zero vector.
Therefore,
\( \vec{b} - \vec{c} = \vec{0} \Rightarrow \vec{b} = \vec{c} \)
Find the Derivative \( \frac{dy}{dx} \)
Given:\[ y = \cos(x^2) + \cos(2x) + \cos^2(x^2) + \cos(x^x) \]
Solve the following LPP graphically: Maximize: \[ Z = 2x + 3y \] Subject to: \[ \begin{aligned} x + 4y &\leq 8 \quad \text{(1)} \\ 2x + 3y &\leq 12 \quad \text{(2)} \\ 3x + y &\leq 9 \quad \text{(3)} \\ x &\geq 0,\quad y \geq 0 \quad \text{(non-negativity constraints)} \end{aligned} \]