7.4 Q-3

Question Statement

Find the area of the triangle formed by the points $P$, $Q$, and $R$.

(i) $P(0,0,0)$, $Q(2,3,2)$, $R(-1,1,4)$

(ii) $P(1,-1,-1)$, $Q(2,0,-1)$, $R(0,2,1)$

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Background and Explanation

To solve for the area of a triangle formed by three points in a 3D space, we use the cross product method. The formula for the area of a triangle with vertices $P$, $Q$, and $R$ is:

$$\text{Area of } \triangle PQR = \frac{1}{2} | \overrightarrow{PQ} \times \overrightarrow{PR} |$$

Where:

• $\overrightarrow{PQ}$ is the vector from point $P$ to point $Q$,
• $\overrightarrow{PR}$ is the vector from point $P$ to point $R$,

The magnitude of the cross product of two vectors gives the area of the parallelogram formed by these vectors, and half of this value gives the area of the triangle.

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Solution

Part (i)

We are given the points $P(0,0,0)$, $Q(2,3,2)$, and $R(-1,1,4)$. The area can be found using the formula:

$$\text{Area} = \frac{1}{2} | \overrightarrow{PQ} \times \overrightarrow{PR} |$$

Step 1: Calculate the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$

$$\overrightarrow{PQ} = Q - P = (2,3,2) - (0,0,0) = (2,3,2)$$ $$\overrightarrow{PR} = R - P = (-1,1,4) - (0,0,0) = (-1,1,4)$$

Step 2: Find the cross product $\overrightarrow{PQ} \times \overrightarrow{PR}$

The cross product formula in component form is:

$$\overrightarrow{PQ} \times \overrightarrow{PR} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 2 & 3 & 2 -1 & 1 & 4 \end{array} \right|$$

Expanding the determinant:

$$= \hat{i} \left( 3 \times 4 - 2 \times 1 \right) - \hat{j} \left( 2 \times 4 - 2 \times (-1) \right) + \hat{k} \left( 2 \times 1 - 3 \times (-1) \right)$$ $$= \hat{i} (12 - 2) - \hat{j} (8 + 2) + \hat{k} (2 + 3)$$ $$= 10\hat{i} - 10\hat{j} + 5\hat{k}$$

Step 3: Calculate the magnitude of the cross product

$$|\overrightarrow{PQ} \times \overrightarrow{PR}| = \sqrt{(10)^2 + (-10)^2 + (5)^2}$$ $$= \sqrt{100 + 100 + 25} = \sqrt{225} = 15$$

Step 4: Calculate the area

$$\text{Area} = \frac{1}{2} \times 15 = \frac{15}{2} , \text{square units}$$

Part (ii)

Now, we are given the points $P(1,-1,-1)$, $Q(2,0,-1)$, and $R(0,2,1)$. The same steps are followed:

Step 1: Calculate the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$

$$\overrightarrow{PQ} = Q - P = (2,0,-1) - (1,-1,-1) = (1,1,0)$$ $$\overrightarrow{PR} = R - P = (0,2,1) - (1,-1,-1) = (-1,3,2)$$

Step 2: Find the cross product $\overrightarrow{PQ} \times \overrightarrow{PR}$

Using the determinant formula:

$$\overrightarrow{PQ} \times \overrightarrow{PR} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 1 & 1 & 0 -1 & 3 & 2 \end{array} \right|$$

Expanding the determinant:

$$= \hat{i} \left( 1 \times 2 - 0 \times 3 \right) - \hat{j} \left( 1 \times 2 - 0 \times (-1) \right) + \hat{k} \left( 1 \times 3 - 1 \times (-1) \right)$$ $$= \hat{i} (2) - \hat{j} (2) + \hat{k} (3 + 1)$$ $$= 2\hat{i} - 2\hat{j} + 4\hat{k}$$

Step 3: Calculate the magnitude of the cross product

$$|\overrightarrow{PQ} \times \overrightarrow{PR}| = \sqrt{(2)^2 + (-2)^2 + (4)^2}$$ $$= \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$$

Step 4: Calculate the area

$$\text{Area} = \frac{1}{2} \times 2\sqrt{6} = \sqrt{6} , \text{square units}$$

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Key Formulas or Methods Used

• Cross Product Formula:

$$\overrightarrow{A} \times \overrightarrow{B} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} A_x & A_y & A_z B_x & B_y & B_z \end{array} \right|$$

• Area of Triangle:

$$\text{Area} = \frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}|$$

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Summary of Steps

1. For Part (i):

   • Calculate vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.
   • Find the cross product $\overrightarrow{PQ} \times \overrightarrow{PR}$.
   • Compute the magnitude of the cross product.
   • Calculate the area as $\frac{1}{2}$ of the magnitude.

2. For Part (ii):

   • Repeat the steps for the given points $P(1,-1,-1)$, $Q(2,0,-1)$, and $R(0,2,1)$.