7.4 Q-4

Question Statement

Find the area of the parallelogram formed by the vertices $A$, $B$, $C$, and $D$.

(i) $A(0,0,0), B(1,2,3), C(2,-1,1), D(3,1,4)$

(ii) $A(1,2,-1), B(4,2,-3), C(6,-5,2), D(9,-5,0)$

(iii) $A(-1,1,1), B(-1,2,2), C(-3,4,-5), D(-3,5,-4)$

---

Background and Explanation

To find the area of a parallelogram in 3D space, we use the following formula:

$$\text{Area of parallelogram} = |\overrightarrow{AB} \times \overrightarrow{AC}|$$

Where:

• $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are vectors from point $A$ to points $B$ and $C$, respectively.
• The cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ gives a vector perpendicular to the parallelogram’s plane, and its magnitude represents the area of the parallelogram.

---

Solution

Part (i)

We are given the points $A(0,0,0)$, $B(1,2,3)$, $C(2,-1,1)$, and $D(3,1,4)$. First, we check which sides are parallel. We find that $\overrightarrow{AB} = \overrightarrow{CD}$, so $ABCD$ is a parallelogram.

Step 1: Calculate the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$

$$\overrightarrow{AB} = B - A = (1,2,3) - (0,0,0) = (1,2,3)$$ $$\overrightarrow{AC} = C - A = (2,-1,1) - (0,0,0) = (2,-1,1)$$

Step 2: Find the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$

Using the determinant formula for the cross product:

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

Expanding the determinant:

$$= \hat{i}(2+3) - \hat{j}(1-6) + \hat{k}(-1-4)$$ $$= 5 \hat{i} + 5 \hat{j} - 5 \hat{k}$$

Step 3: Calculate the magnitude of the cross product

$$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(5)^2 + (5)^2 + (-5)^2}$$ $$= \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3}$$

Step 4: Calculate the area

$$\text{Area} = 5\sqrt{3} , \text{square units}$$

---

Part (ii)

We are given the points $A(1,2,-1)$, $B(4,2,-3)$, $C(6,-5,2)$, and $D(9,-5,0)$. We find that $\overrightarrow{AB} = \overrightarrow{CD}$, so $ABCD$ is a parallelogram.

Step 1: Calculate the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$

$$\overrightarrow{AB} = B - A = (4,2,-3) - (1,2,-1) = (3,0,-2)$$ $$\overrightarrow{AC} = C - A = (6,-5,2) - (1,2,-1) = (5,-7,3)$$

Step 2: Find the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$

Using the determinant formula for the cross product:

$$\overrightarrow{AB} \times \overrightarrow{AC} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 3 & 0 & -2 5 & -7 & 3 \end{array} \right|$$

Expanding the determinant:

$$= \hat{i}(0 - (-14)) - \hat{j}(9 + 10) + \hat{k}(-21 - 0)$$ $$= -14 \hat{i} - 19 \hat{j} - 21 \hat{k}$$

Step 3: Calculate the magnitude of the cross product

$$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-14)^2 + (-19)^2 + (-21)^2}$$ $$= \sqrt{196 + 361 + 441} = \sqrt{998}$$

Step 4: Calculate the area

$$\text{Area} = \sqrt{998} , \text{square units}$$

---

Part (iii)

We are given the points $A(-1,1,1)$, $B(-1,2,2)$, $C(-3,4,-5)$, and $D(-3,5,-4)$. We find that $\overrightarrow{AB} = \overrightarrow{CD}$, so $ABCD$ is a parallelogram.

Step 1: Calculate the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$

$$\overrightarrow{AB} = B - A = (-1,2,2) - (-1,1,1) = (0,1,1)$$ $$\overrightarrow{AC} = C - A = (-3,4,-5) - (-1,1,1) = (-2,3,-6)$$

Step 2: Find the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$

Using the determinant formula for the cross product:

$$\overrightarrow{AB} \times \overrightarrow{AC} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} 0 & 1 & 1 -2 & 3 & -6 \end{array} \right|$$

Expanding the determinant:

$$= \hat{i}(-6 - 3) - \hat{j}(0 + 2) + \hat{k}(0 + 2)$$ $$= -9 \hat{i} - 2 \hat{j} + 2 \hat{k}$$

Step 3: Calculate the magnitude of the cross product

$$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-9)^2 + (-2)^2 + (2)^2}$$ $$= \sqrt{81 + 4 + 4} = \sqrt{89}$$

Step 4: Calculate the area

$$\text{Area} = \sqrt{89} , \text{square units}$$

---

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 Parallelogram:

$$\text{Area} = |\overrightarrow{AB} \times \overrightarrow{AC}|$$

---

Summary of Steps

1. For each part:
   • Calculate the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$.
   • Find the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$.
   • Calculate the magnitude of the cross product.
   • The magnitude gives the area of the parallelogram.