Let's analyze the vector BC based on the given image, where we need to:

1. Represent BC in column vector form.
2. Express BC in terms of unit vectors $\mathbf{i}$ and $\mathbf{j}$.
3. Calculate the magnitude (length) of vector BC.

Solution:

From the graph, we observe:

• The coordinates of point $B$ are $(3, 2)$.
• The coordinates of point $C$ are $(7, 4)$.

Step-by-Step Solution:

(a) Column Vector Form of BC

The column vector form for a vector BC from point $B(x_1, y_1)$ to $C(x_2, y_2)$ is: $\mathbf{BC} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$ Using $B(3, 2)$ and $C(7, 4)$: $\mathbf{BC} = \begin{pmatrix} 7 - 3 \\ 4 - 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$

(b) Unit Vector Form $\mathbf{i}$ and $\mathbf{j}$

Expressing BC in terms of unit vectors $\mathbf{i}$ (horizontal) and $\mathbf{j}$ (vertical): $\mathbf{BC} = 4\mathbf{i} + 2\mathbf{j}$

(c) Magnitude of Vector BC

The magnitude $|\mathbf{BC}|$ of the vector BC is calculated by: $|\mathbf{BC}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Substituting the values: $|\mathbf{BC}| = \sqrt{(7 - 3)^2 + (4 - 2)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \approx 4.47$

Summary of Answers:

• (a) Column vector form: $\begin{pmatrix} 4 \\ 2 \end{pmatrix}$
• (b) Unit vector form: $4\mathbf{i} + 2\mathbf{j}$
• (c) Magnitude of vector BC: $2\sqrt{5} \approx 4.47$

Would you like additional details on any part of the solution?

Related Questions

1. How do you calculate the midpoint of a vector given its endpoints?
2. What is the direction angle of vector BC with respect to the x-axis?
3. How do you find the unit vector in the direction of vector BC?
4. How would the solution change if the vector pointed from $C$ to $B$ instead?
5. What is the general formula for converting between vector forms?

Tip

When calculating vector magnitudes, remember to square each component, add them, and then take the square root for the final result.