In a triangle ABC , right angled at the vertex A , if the position vectors of A , B and C are respectively `3 hati + hatj-hatk , -hati + 3hatj + phatk` and `5 hati + q hatj - 4 hatk` , then the point (p,q) lies on a line

Options

A. making an obtuse angle which the positive direction of x-axis.
B. parallel to x-axis.
C. parallel to y-axis.
D. making an acute angle with the positive directions of x-axis.

Answer

To solve the given problem, we need to analyze the position vectors of points A, B, and C in triangle ABC, which is right-angled at vertex A. The position vectors are given as follows:

- Position vector of A: \(\vec{A} = 3\hat{i} + \hat{j} - \hat{k}\)
- Position vector of B: \(\vec{B} = -\hat{i} + 3\hat{j} + p\hat{k}\)
- Position vector of C: \(\vec{C} = 5\hat{i} + q\hat{j} - 4\hat{k}\)

### Step 1: Find the vectors AB and AC
We first find the vectors \(\vec{AB}\) and \(\vec{AC}\):

\[ \vec{AB} = \vec{B} - \vec{A} = (-\hat{i} + 3\hat{j} + p\hat{k}) - (3\hat{i} + \hat{j} - \hat{k}) \]

Calculating this gives:

\[ \vec{AB} = (-4\hat{i} + 2\hat{j} + (p + 1)\hat{k}) \]

Next, we find \(\vec{AC}\):

\[ \vec{AC} = \vec{C} - \vec{A} = (5\hat{i} + q\hat{j} - 4\hat{k}) - (3\hat{i} + \hat{j} - \hat{k}) \]

Calculating this gives:

\[ \vec{AC} = (2\hat{i} + (q - 1)\hat{j} - 3\hat{k}) \]

### Step 2: Use the property of right-angled triangles
Since triangle ABC is right-angled at A, the vectors \(\vec{AB}\) and \(\vec{AC}\) must be perpendicular. This means their dot product must equal zero:

\[ \vec{AB} \cdot \vec{AC} = 0 \]

Calculating the dot product:

\[ (-4\hat{i} + 2\hat{j} + (p + 1)\hat{k}) \cdot (2\hat{i} + (q - 1)\hat{j} - 3\hat{k}) = 0 \]

Calculating each component of the dot product:

\[ (-4)(2) + (2)(q - 1) + (p + 1)(-3) = 0 \]

This simplifies to:

\[ -8 + 2(q - 1) - 3(p + 1) = 0 \]

Expanding this gives:

\[ -8 + 2q - 2 - 3p - 3 = 0 \]

Combining like terms results in:

\[ -3p + 2q - 13 = 0 \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:

\[ 3p - 2q + 13 = 0 \]

### Step 4: Identify the line equation
This equation can be rewritten in the standard form of a line:

\[ 3p - 2q = -13 \]

This represents a line in the \(pq\)-plane.

### Conclusion
The point \((p, q)\) lies on the line defined by the equation \(3p - 2q + 13 = 0\).