The coordinates of the vertices P,Q and R of a triangle PQR are `(1,-1,1),(3,-2,2) and (0,2,6)` respectively. If `angleRQP=theta`, then what is `anglePRQ` equal to ?

Options

A. `30^@+theta`
B. `45^@-theta`
C. `60^@-theta`
D. `90^@-theta`

Answer

To find the angle PRQ given that angle RQP = θ, we can follow these steps:

### Step 1: Identify the coordinates of the points
The coordinates of the vertices of triangle PQR are:
- P(1, -1, 1)
- Q(3, -2, 2)
- R(0, 2, 6)

### Step 2: Calculate the direction ratios of PQ and PR
To find the direction ratios, we need to calculate the vectors PQ and PR.

1. **Vector PQ**:
\[ PQ = Q - P = (3 - 1, -2 - (-1), 2 - 1) = (2, -1, 1) \]

2. **Vector PR**:
\[ PR = R - P = (0 - 1, 2 - (-1), 6 - 1) = (-1, 3, 5) \]

### Step 3: Check if PQ is perpendicular to PR
To check if the vectors PQ and PR are perpendicular, we calculate the dot product of PQ and PR. If the dot product is zero, the vectors are perpendicular.

\[ PQ \cdot PR = (2)(-1) + (-1)(3) + (1)(5) \]
Calculating this:
\[ = -2 - 3 + 5 = 0 \]

Since the dot product is zero, we conclude that PQ is perpendicular to PR.

### Step 4: Relate the angles
Since PQ is perpendicular to PR, we have:
\[ \angle QPR = 90^\circ \]

Given that:
\[ \angle RQP = \theta \]

Using the triangle angle sum property, we know that:
\[ \angle QPR + \angle RQP + \angle PRQ = 180^\circ \]

Substituting the known angles:
\[ 90^\circ + \theta + \angle PRQ = 180^\circ \]

### Step 5: Solve for angle PRQ
Rearranging the equation gives:
\[ \angle PRQ = 180^\circ - 90^\circ - \theta \]
\[ \angle PRQ = 90^\circ - \theta \]

### Final Answer
Thus, the angle PRQ is equal to \(90^\circ - \theta\).