Question

In each case, find the point \(Q:\) a. \(\overrightarrow{P Q}=\left[\begin{array}{r}2 \\ 0 \\\ -3\end{array}\right]\) and \(P=P(2,-3,1)\) b. \(\overrightarrow{P Q}=\left[\begin{array}{r}-1 \\ 4 \\\ 7\end{array}\right]\) and \(P=P(1,3,-4)\)

Short answer

a. \((4, -3, -2)\), b. \((0, 7, 3)\)

Step-by-step solution

Understanding the Vector Representation

The vector \( \overrightarrow{PQ} \) is given by the difference between point \( Q \) and point \( P \). If \( Q = (x, y, z) \), then \( \overrightarrow{PQ} = [x- x_P, y - y_P, z - z_P] \). We need to use this formula to calculate the coordinates of \( Q \).

Solving Part (a)

For part (a), we have \( P = (2, -3, 1) \) and \( \overrightarrow{PQ} = \begin{bmatrix} 2 \ 0 \ -3 \end{bmatrix} \). The calculation is:- \( x_Q = x_P + 2 = 2 + 2 = 4 \) - \( y_Q = y_P + 0 = -3 + 0 = -3 \) - \( z_Q = z_P - 3 = 1 - 3 = -2 \) Thus, \( Q = (4, -3, -2) \).

Solving Part (b)

For part (b), we have \( P = (1, 3, -4) \) and \( \overrightarrow{PQ} = \begin{bmatrix} -1 \ 4 \ 7 \end{bmatrix} \). The calculation is:- \( x_Q = x_P - 1 = 1 - 1 = 0 \) - \( y_Q = y_P + 4 = 3 + 4 = 7 \) - \( z_Q = z_P + 7 = -4 + 7 = 3 \) Thus, \( Q = (0, 7, 3) \).

Key concepts

Understanding Vector Representation

Vector representation is a powerful tool in mathematics and physics to describe quantities having direction as well as magnitude. In this context, vectors help us understand how to translate from one point in space to another. A vector can be visually imagined as an arrow pointing from one point to another. The direction of the arrow indicates the direction of movement, and its length indicates the distance between the points.

When we say vector \( \overrightarrow{PQ} \), it tells us that it starts at point \( P \) and ends at point \( Q \). The components of the vector are the differences in the coordinates between point \( Q \) and point \( P \). For instance, if the given vector is \( \overrightarrow{PQ} = \begin{bmatrix} a \ b \ c \end{bmatrix} \), this indicates how far along each axis the point \( Q \) is from point \( P \).

Remember:

• Vectors have both magnitude and direction.
• Vector \( \overrightarrow{PQ} \) is given as \( \begin{bmatrix} x_Q - x_P \ y_Q - y_P \ z_Q - z_P \end{bmatrix} \).

Decoding Coordinates Calculation

Calculating the coordinates of a point using vectors involves understanding the relationship between the starting and ending points. Suppose the coordinates of point \( P \) are \( (x_P, y_P, z_P) \), and the vector \( \overrightarrow{PQ} \) that translates \( P \) to \( Q \) is given. Then we can find the coordinates of point \( Q \) using simple arithmetic.

- **Step 1:** Take the x-coordinate of point \( P \). Add or subtract the x-component of the vector to find \( x_Q \).
- **Step 2:** Do the same for the y-coordinate. Here, use the y-component of the vector.
- **Step 3:** Finally, calculate the z-coordinate using the z-component of the vector.

### Example:
Let's find point \( Q \) when \( P = (2, -3, 1) \) and \( \overrightarrow{PQ} = \begin{bmatrix} 2 \ 0 \ -3 \end{bmatrix} \):

• \( x_Q = 2 + 2 = 4 \)
• \( y_Q = -3 + 0 = -3 \)
• \( z_Q = 1 - 3 = -2 \)

This results in \( Q \) being located at \( (4, -3, -2) \). It’s all about understanding the translation in each coordinate direction.

Point Translation in Space

Point translation using vectors is essentially about moving points across a grid or space. This concept allows us to shift a point by a specific amount along three-dimensional axes, using a vector's defined components. Each component of the vector defines how far and in what direction the point will move along each axis.

### How It Works: - The **x-component** tells how much to move left or right. - The **y-component** guides the movement up or down. - The **z-component** affects the movement forward or backward in space.

By translating a point \( P = (x_P, y_P, z_P) \) using vector \( \overrightarrow{PQ} = \begin{bmatrix} a \ b \ c \end{bmatrix} \), we compute \( Q \) by applying: - \( x_Q = x_P + a \)- \( y_Q = y_P + b \)- \( z_Q = z_P + c \)

### Practical Application:If you want to move \( P = (1, 3, -4) \) via vector \( \overrightarrow{PQ} = \begin{bmatrix} -1 \ 4 \ 7 \end{bmatrix} \):

• Slide one unit left: \( x_Q = 1 - 1 = 0 \)
• Raise four units up: \( y_Q = 3 + 4 = 7 \)
• Advance seven units forward: \( z_Q = -4 + 7 = 3 \)

This gives you the new point \( Q = (0, 7, 3) \). The essence is to adjust your initial point by the vector to find the new location.