Two lines are perpendicular if their direction vectors are also perpendicular. This means that the dot product of their direction vectors must be zero.
Step 1: Determine the direction vectors. The direction vector of \(\vec{r}_1\) is: \[ \vec{d}_1 = 2\hat{i} + p\hat{j} + 5\hat{k}. \] The direction vector of \(\vec{r}_2\) is: \[ \vec{d}_2 = 3\hat{i} - p\hat{j} + p\hat{k}. \]
Step 2: Calculate the dot product. The dot product of \(\vec{d}_1\) and \(\vec{d}_2\) is: \[ \vec{d}_1 \cdot \vec{d}_2 = (2)(3) + (p)(-p) + (5)(p). \] Simplifying the expression: \[ \vec{d}_1 \cdot \vec{d}_2 = 6 - p^2 + 5p. \]
Step 3: Set the dot product equal to zero. Since the lines are perpendicular, the dot product must be zero: \[ 6 - p^2 + 5p = 0. \]
Step 4: Solve the quadratic equation. Rearrange the terms: \[ p^2 - 5p - 6 = 0. \] Factor the quadratic expression: \[ (p - 6)(p + 1) = 0. \] Thus, the possible solutions for \( p \) are: \[ p = 6 \quad \text{or} \quad p = -1. \]
Step 5: Confirm the solution. Both \( p = 6 \) and \( p = -1 \) satisfy the condition of perpendicularity. In this case, the solution required is \( p = 6 \). Final Answer: \[ \boxed{6} \]
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \( f(x + y) = f(x) f(y) \) for all \( x, y \in \mathbb{R} \). If \( f'(0) = 4a \) and \( f \) satisfies \( f''(x) - 3a f'(x) - f(x) = 0 \), where \( a > 0 \), then the area of the region R = {(x, y) | 0 \(\leq\) y \(\leq\) f(ax), 0 \(\leq\) x \(\leq\) 2\ is :