7.2 Q-9

Question Statement

The position vectors of the points $A$, $B$, $C$, and $D$ are:

• $A = 2 \underline{i} - \underline{j} + \underline{k}$
• $B = 3 \underline{i} + \underline{j}$
• $C = 2 \underline{i} + 4 \underline{j} - 2 \underline{k}$
• $D = -\underline{i} - 2 \underline{j} + \underline{k}$

Show that the vector $\mathbf{AB}$ is parallel to the vector $\mathbf{CD}$.

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Background and Explanation

To solve this problem, we need to:

1. Find the vector $\mathbf{AB}$ by subtracting the position vector of $A$ from the position vector of $B$.
2. Find the vector $\mathbf{CD}$ by subtracting the position vector of $C$ from the position vector of $D$.
3. Check if the vectors are parallel by verifying if one vector is a scalar multiple of the other.

Two vectors are parallel if one vector is a scalar multiple of the other. In other words, if: $\mathbf{AB} = k \mathbf{CD}$ for some scalar $k$, then the vectors are parallel.

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Solution

Step 1: Find the vector $\mathbf{AB}$

The vector $\mathbf{AB}$ is calculated by subtracting the position vector of $A$ from the position vector of $B$:

$$\mathbf{AB} = \mathbf{B} - \mathbf{A} = (3 \underline{i} + \underline{j}) - (2 \underline{i} - \underline{j} + \underline{k})$$

Simplifying:

$$\mathbf{AB} = (3 - 2) \underline{i} + (1 - (-1)) \underline{j} + (0 - 1) \underline{k}$$ $$\mathbf{AB} = \underline{i} + 2 \underline{j} - \underline{k} \tag{1}$$

Step 2: Find the vector $\mathbf{CD}$

Next, calculate the vector $\mathbf{CD}$ by subtracting the position vector of $C$ from the position vector of $D$:

$$\mathbf{CD} = \mathbf{D} - \mathbf{C} = (-\underline{i} - 2 \underline{j} + \underline{k}) - (2 \underline{i} + 4 \underline{j} - 2 \underline{k})$$

Simplifying:

$$\mathbf{CD} = (-1 - 2) \underline{i} + (-2 - 4) \underline{j} + (1 - (-2)) \underline{k}$$ $$\mathbf{CD} = -3 \underline{i} - 6 \underline{j} + 3 \underline{k}$$

Step 3: Check if $\mathbf{AB}$ and $\mathbf{CD}$ are parallel

To check if the vectors $\mathbf{AB}$ and $\mathbf{CD}$ are parallel, we see if $\mathbf{CD}$ is a scalar multiple of $\mathbf{AB}$.

From equation (1), we know that: $\mathbf{AB} = \underline{i} + 2 \underline{j} - \underline{k}$

We can factor out $-3$ from $\mathbf{CD}$:

$$\mathbf{CD} = -3(\underline{i} + 2 \underline{j} - \underline{k})$$

This shows that:

$$\mathbf{CD} = -3 \mathbf{AB}$$

Since $\mathbf{CD}$ is a scalar multiple of $\mathbf{AB}$, the vectors $\mathbf{AB}$ and $\mathbf{CD}$ are parallel, but they point in opposite directions.

Thus, $\mathbf{AB}$ is parallel to $\mathbf{CD}$, but they are in opposite directions.

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Key Formulas or Methods Used

• Vector Subtraction: To find the vector between two points, subtract the position vector of the initial point from the position vector of the final point. $\mathbf{AB} = \mathbf{B} - \mathbf{A}$

• Scalar Multiple: Two vectors are parallel if one is a scalar multiple of the other. $\mathbf{AB} = k \mathbf{CD}$

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Summary of Steps

1. Find $\mathbf{AB}$ by subtracting the position vectors of $A$ and $B$.
2. Find $\mathbf{CD}$ by subtracting the position vectors of $C$ and $D$.
3. Check parallelism by verifying if $\mathbf{CD}$ is a scalar multiple of $\mathbf{AB}$.
4. Conclude that $\mathbf{AB}$ is parallel to $\mathbf{CD}$, but they are in opposite directions.