Key Concepts and Formulas

• Position Vectors: A position vector represents the location of a point in space relative to an origin. If $\vec{r}_A$ and $\vec{r}_B$ are the position vectors of points A and B, then the vector $\vec{AB}$ is given by $\vec{AB} = \vec{r}_B - \vec{r}_A$.
• Dot Product of Perpendicular Vectors: Two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular (orthogonal) if and only if their dot product is zero, i.e., $\vec{u} \cdot \vec{v} = 0$.
• Dot Product Calculation: For vectors $\vec{u} = u_x\widehat i + u_y\widehat j + u_z\widehat k$ and $\vec{v} = v_x\widehat i + v_y\widehat j + v_z\widehat k$, their dot product is $\vec{u} \cdot \vec{v} = u_xv_x + u_yv_y + u_zv_z$.
• Slope of a Line: For a linear equation $Ax + By + C = 0$, the slope is $m = -\frac{A}{B}$. The angle $\theta$ the line makes with the positive x-axis is given by $\tan \theta = m$.
  • If $m > 0$, $\theta$ is acute ($0^\circ < \theta < 90^\circ$).
  • If $m < 0$, $\theta$ is obtuse ($90^\circ < \theta < 180^\circ$).
  • If $m = 0$, the line is parallel to the x-axis.
  • If the slope is undefined (i.e., $B=0$), the line is parallel to the y-axis.

Step-by-Step Solution

1. Determine the vectors representing the sides $\vec{AB}$ and $\vec{AC}$: We are given the position vectors of vertices A, B, and C: $\vec{r}_A = 3\widehat i + \widehat j - \widehat k$ $\vec{r}_B = -\widehat i + 3\widehat j + p\widehat k$ $\vec{r}_C = 5\widehat i + q\widehat j - 4\widehat k$

   The vector $\vec{AB}$ is found by subtracting the position vector of A from the position vector of B: $\vec{AB} = \vec{r}_B - \vec{r}_A = (-\widehat i + 3\widehat j + p\widehat k) - (3\widehat i + \widehat j - \widehat k)$ $\vec{AB} = (-1 - 3)\widehat i + (3 - 1)\widehat j + (p - (-1))\widehat k$ $\vec{AB} = -4\widehat i + 2\widehat j + (p+1)\widehat k$

   The vector $\vec{AC}$ is found by subtracting the position vector of A from the position vector of C: $\vec{AC} = \vec{r}_C - \vec{r}_A = (5\widehat i + q\widehat j - 4\widehat k) - (3\widehat i + \widehat j - \widehat k)$ $\vec{AC} = (5 - 3)\widehat i + (q - 1)\widehat j + (-4 - (-1))\widehat k$ $\vec{AC} = 2\widehat i + (q-1)\widehat j - 3\widehat k$

2. Apply the condition for a right-angled triangle at A: Since the triangle ABC is right-angled at vertex A, the vectors $\vec{AB}$ and $\vec{AC}$ must be perpendicular. This means their dot product is zero: $\vec{AB} \cdot \vec{AC} = 0$

3. Calculate the dot product and form an equation in terms of p and q: Using the components of $\vec{AB}$ and $\vec{AC}$: $(-4)(2) + (2)(q-1) + (p+1)(-3) = 0$ Now, we simplify this equation: $-8 + 2q - 2 - 3p - 3 = 0$ Combine the constant terms: $-3p + 2q - 13 = 0$

4. Rearrange the equation to represent a line and determine its slope: The equation $-3p + 2q - 13 = 0$ describes the relationship between $p$ and $q$. This is the equation of a straight line in the $pq$-plane. To analyze the line's orientation, we can rewrite it in the form $Ap + Bq + C = 0$, which is already done. The slope of this line, considering $p$ as the x-axis and $q$ as the y-axis, is given by $m = -\frac{A}{B}$. Here, $A = -3$ and $B = 2$. $m = -\frac{-3}{2} = \frac{3}{2}$

5. Interpret the slope to determine the line's orientation: The slope of the line is $m = \frac{3}{2}$. Since the slope is positive ($m > 0$), the line makes an acute angle with the positive direction of the $p$-axis (which corresponds to the x-axis in the problem statement's context of $(p, q)$ lying on a line).

Common Mistakes & Tips

• Order of Subtraction for Vectors: Always subtract the position vector of the initial point from the position vector of the terminal point (e.g., $\vec{AB} = \vec{r}_B - \vec{r}_A$).
• Dot Product Calculation Errors: Be careful with signs when multiplying corresponding components and summing them up.
• Interpreting the Slope: Remember that a positive slope indicates an acute angle, a negative slope indicates an obtuse angle, a zero slope indicates a horizontal line (parallel to the x-axis), and an undefined slope indicates a vertical line (parallel to the y-axis).

Summary

The problem requires us to find the locus of the point $(p, q)$ given that a triangle with vertices A, B, and C is right-angled at A. We used the property that the dot product of the vectors representing the sides $\vec{AB}$ and $\vec{AC}$ must be zero. After calculating these vectors and their dot product, we derived a linear equation relating $p$ and $q$. Analyzing the slope of this linear equation, we found it to be positive, which means the point $(p, q)$ lies on a line that makes an acute angle with the positive x-axis.

The final answer is $\boxed{A}$.