Question

In each case, find a point \(Q\) such that \(\overrightarrow{P Q}\) has (i) the same direction as \(\mathbf{v}\); (ii) the opposite direction to \(\mathbf{v}\). a. \(P(-1,2,2), \mathbf{v}=\left[\begin{array}{l}1 \\ 3 \\\ 1\end{array}\right]\) b. \(P(3,0,-1), \mathbf{v}=\left[\begin{array}{r}2 \\ -1 \\\ 3\end{array}\right]\)

Short answer

a. Q (same direction): (0, 5, 3), (opposite): (-2, -1, 1) b. Q (same direction): (5, -1, 2), (opposite): (1, 1, -4)

Step-by-step solution

Understanding the Problem

We need to find a point \( Q \) such that the vector \( \overrightarrow{PQ} \) has either the same direction or the opposite direction as the given vector \( \mathbf{v} \). This can be achieved by setting \( \overrightarrow{PQ} = k\mathbf{v} \) for some scalar \( k \). For the same direction, \( k > 0 \); for the opposite direction, \( k < 0 \).

Calculate Q for Same Direction (Part a)

To have the same direction as \( \mathbf{v} = \begin{bmatrix} 1 \ 3 \ 1 \end{bmatrix} \), choose \( k = 1 \). Then \( \overrightarrow{PQ} = \begin{bmatrix} Q_x - (-1) \ Q_y - 2 \ Q_z - 2 \end{bmatrix} = \begin{bmatrix} 1 \ 3 \ 1 \end{bmatrix} \). Solving gives \( Q_x = 0 \), \( Q_y = 5 \), \( Q_z = 3 \). Thus, \( Q = (0, 5, 3) \).

Calculate Q for Opposite Direction (Part a)

Choose \( k = -1 \) for the opposite direction. Then \( \overrightarrow{PQ} = \begin{bmatrix} Q_x - (-1) \ Q_y - 2 \ Q_z - 2 \end{bmatrix} = \begin{bmatrix} -1 \ -3 \ -1 \end{bmatrix} \). Solving gives \( Q_x = -2 \), \( Q_y = -1 \), \( Q_z = 1 \). Thus, \( Q = (-2, -1, 1) \).

Calculate Q for Same Direction (Part b)

To have the same direction as \( \mathbf{v} = \begin{bmatrix} 2 \ -1 \ 3 \end{bmatrix} \), choose \( k = 1 \). Then \( \overrightarrow{PQ} = \begin{bmatrix} Q_x - 3 \ Q_y - 0 \ Q_z + 1 \end{bmatrix} = \begin{bmatrix} 2 \ -1 \ 3 \end{bmatrix} \). Solving gives \( Q_x = 5 \), \( Q_y = -1 \), \( Q_z = 2 \). Thus, \( Q = (5, -1, 2) \).

Calculate Q for Opposite Direction (Part b)

Choose \( k = -1 \) for the opposite direction. Then \( \overrightarrow{PQ} = \begin{bmatrix} Q_x - 3 \ Q_y - 0 \ Q_z + 1 \end{bmatrix} = \begin{bmatrix} -2 \ 1 \ -3 \end{bmatrix} \). Solving gives \( Q_x = 1 \), \( Q_y = 1 \), \( Q_z = -4 \). Thus, \( Q = (1, 1, -4) \).

Key concepts

Vector Direction

Understanding vector direction is crucial in vector calculus. It defines the specific way or path along which something moves or points.
Vectors are multidimensional, consisting of both magnitude and direction. Imagine an arrow; the length represents magnitude, while the arrowhead points in the direction of movement. When we talk about the direction of a vector, it always points from the initial point towards the terminal point.
For instance, in the exercise, given the vector \( \mathbf{v} = \begin{bmatrix} 1 \ 3 \ 1 \end{bmatrix} \), we determine the direction by looking at changes across each component:

• The X-component directs one unit in the positive x-direction.
• The Y-component directs three units in the positive y-direction.
• The Z-component directs one unit in the positive z-direction.

This consistent directional component helps us define if another vector has the same or different direction. Calculating this accurately means knowing how far and in what proportion each part of the vector stretches in a coordinate grid.

Scalar Multiplication

Scalar multiplication is fundamental in scaling a vector without altering its direction. When you multiply a vector by a scalar, it effectively stretches or shrinks the vector.
You can visualize this with the scalar \( k \). Imagine the original vector is like a stick, and multiplying by different scalars changes its length while keeping it pointed the same way:

• If \( k > 1 \), the vector stretches, making it longer.
• If \( 0 < k < 1 \), the vector shrinks but remains in the same direction.
• If \( k = 1 \), the vector remains unchanged.

In the exercise, using scalar multiplication helps ensure that \( \overrightarrow{PQ} \) aligns with vector \( \mathbf{v} \). For example, in Part (a), selecting \( k = 1 \) confirms \( \overrightarrow{PQ} = \mathbf{v} = \begin{bmatrix} 1 \ 3 \ 1 \end{bmatrix} \), preserving direction and affirming both point in the same way. Adjusting the scalar helps us set vectors in different orientations for different scenarios.

Opposite Direction

Vectors in opposite directions point towards the exact opposite location with the same magnitude but different orientation.
In mathematical terms, this is achieved by multiplying the vector by a negative scalar. Consider a vector \( \mathbf{v} \); multiplying by \( -1 \) reflects it through the origin, flipping its direction 180 degrees.
This concept is applied in the exercise by choosing \( k = -1 \). For instance, in Part (a), multiplying the direction vector \( \mathbf{v} = \begin{bmatrix} 1 \ 3 \ 1 \end{bmatrix} \) by \( -1 \) results in:

• The X-component becomes \( -1 \), moving it backward relative to the positive X-direction.
• The Y-component becomes \( -3 \), reversing its Y-direction.
• The Z-component becomes \( -1 \), pulling it back along the Z-axis.

Thus, when \( \overrightarrow{PQ} = \begin{bmatrix} -1 \ -3 \ -1 \end{bmatrix} \), it signifies a full reversal in direction, proving vital whenever you need to identify or construct vectors pointing the opposite way. Understanding these vector transformations through scalar and direction manipulation is essential for successful vector calculus.