7.2 Q-1

Question Statement

Given points $A(2,5)$, $B(-1,1)$, and $C(2,-6)$, the following calculations need to be performed:

1. (i) Find $AB$ (the vector from point A to point B)
2. (ii) Calculate $2 \mathrm{A} \vec{B} - C \vec{B}$
3. (iii) Find $2 C \vec{B} - 2 C \vec{A}$

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Background and Explanation

To solve these problems, we need to apply basic vector operations such as subtraction, scalar multiplication, and understanding the relationship between points and vectors.

• Vector Notation: A vector between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ can be calculated as: $\overrightarrow{PQ} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j}$

• Scalar Multiplication: To multiply a vector by a scalar, multiply each of its components by the scalar.

These concepts are essential for finding the required vectors and performing the operations in the problem.

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Solution

Let’s solve each part step by step.

(i) Find AB

The vector from point $A(2, 5)$ to point $B(-1, 1)$ is calculated by subtracting the coordinates of $A$ from those of $B$:

$$\overrightarrow{AB} = B - A = (-1, 1) - (2, 5)$$

Now subtract the corresponding components:

$$\overrightarrow{AB} = (-1 - 2, 1 - 5) = (-3, -4)$$

So, the vector $\overrightarrow{AB} = -3 \mathbf{i} - 4 \mathbf{j}$.

(ii) Calculate $2 A \vec{B} - C \vec{B}$

First, we need to find the vector $\overrightarrow{AB}$ and $\overrightarrow{CB}$:

$$\overrightarrow{AB} = (-1, 1) - (2, 5) = (-3, -4)$$ $$\overrightarrow{CB} = (-1, 1) - (2, -6) = (-3, 7)$$

Now, let’s calculate $2 \overrightarrow{AB} - \overrightarrow{CB}$:

$$2 \overrightarrow{AB} = 2(-3, -4) = (-6, -8)$$

Now subtract $\overrightarrow{CB}$ from $2 \overrightarrow{AB}$:

$$2 \overrightarrow{AB} - \overrightarrow{CB} = (-6, -8) - (-3, 7)$$

This simplifies to:

$$(-6 + 3, -8 - 7) = (-3, -15)$$

So, the result is $-3 \mathbf{i} - 15 \mathbf{j}$.

(iii) Find $2 C \vec{B} - 2 C \vec{A}$

Let’s first calculate the vectors $\overrightarrow{CB}$ and $\overrightarrow{CA}$:

$$\overrightarrow{CB} = (-1, 1) - (2, -6) = (-3, 7)$$ $$\overrightarrow{CA} = (2, 5) - (2, -6) = (0, 11)$$

Now calculate $2 \overrightarrow{CB} - 2 \overrightarrow{CA}$:

$$2 \overrightarrow{CB} = 2(-3, 7) = (-6, 14)$$ $$2 \overrightarrow{CA} = 2(0, 11) = (0, 22)$$

Finally, subtract $2 \overrightarrow{CA}$ from $2 \overrightarrow{CB}$:

$$2 \overrightarrow{CB} - 2 \overrightarrow{CA} = (-6, 14) - (0, 22)$$

This simplifies to:

$$(-6 - 0, 14 - 22) = (-6, -8)$$

So the result is $-6 \mathbf{i} + 8 \mathbf{j}$.

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Key Formulas or Methods Used

• Vector Subtraction: $\overrightarrow{PQ} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j}$

• Scalar Multiplication: $k \overrightarrow{PQ} = (k \cdot (x_2 - x_1), k \cdot (y_2 - y_1))$

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Summary of Steps

1. Find AB: Subtract the coordinates of point $A$ from point $B$ to get the vector $\overrightarrow{AB}$.
2. Calculate $2 A \vec{B} - C \vec{B}$: Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{CB}$, multiply $\overrightarrow{AB}$ by 2, then subtract $\overrightarrow{CB}$.
3. Find $2 C \vec{B} - 2 C \vec{A}$: Find the vectors $\overrightarrow{CB}$ and $\overrightarrow{CA}$, multiply each by 2, then subtract the results.