Question

Find all vectors \(\mathbf{u}\) that are parallel to \(\mathbf{v}=\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right]\) and satisfy \(\|\mathbf{u}\|=3\|\mathbf{v}\| .\)

Short answer

Vectors are \( \mathbf{u} = \begin{bmatrix} 9 \\ -6 \\ 3 \end{bmatrix} \) and \( \mathbf{u} = \begin{bmatrix} -9 \\ 6 \\ -3 \end{bmatrix} \).

Step-by-step solution

Determine the magnitude of \( \mathbf{v} \)

To find the magnitude of the vector \( \mathbf{v} = \begin{bmatrix} 3 \ -2 \ 1 \end{bmatrix} \), we use the formula for the magnitude of a vector, which is \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \). Calculate: \( \|\mathbf{v}\| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \).

Calculate the magnitude of \( \mathbf{u} \)

The problem states that the magnitude of \( \mathbf{u} \) is three times the magnitude of \( \mathbf{v} \). Therefore, \( \|\mathbf{u}\| = 3\|\mathbf{v}\| = 3 \times \sqrt{14} = 3\sqrt{14} \).

Express \( \mathbf{u} \) in terms of \( \mathbf{v} \)

Vectors are parallel if one is a scalar multiple of the other. Thus, we can express \( \mathbf{u} \) as \( \mathbf{u} = k\mathbf{v} \), where \( k \) is some scalar.

Set up the equation for the magnitude of \( \mathbf{u} \)

Since \( \mathbf{u} = k\mathbf{v} \), we have \( \|\mathbf{u}\| = |k|\|\mathbf{v}\| \). Substituting \( \|\mathbf{u}\| = 3\sqrt{14} \) and \( \|\mathbf{v}\| = \sqrt{14} \) from previous steps, we get \( |k|\sqrt{14} = 3\sqrt{14} \).

Solve for \( k \)

From the equation \( |k|\sqrt{14} = 3\sqrt{14} \), we can divide both sides by \( \sqrt{14} \) to get \( |k| = 3 \). Therefore, \( k = 3 \) or \( k = -3 \).

Find the vectors \( \mathbf{u} \)

Using the values of \( k \) from Step 5, we find \( \mathbf{u} \) as follows: for \( k = 3 \), \( \mathbf{u} = 3\mathbf{v} = 3\begin{bmatrix} 3 \ -2 \ 1 \end{bmatrix} = \begin{bmatrix} 9 \ -6 \ 3 \end{bmatrix} \). For \( k = -3 \), \( \mathbf{u} = -3\mathbf{v} = -3\begin{bmatrix} 3 \ -2 \ 1 \end{bmatrix} = \begin{bmatrix} -9 \ 6 \ -3 \end{bmatrix} \).

Key concepts

Vector Magnitude

Understanding the concept of vector magnitude is essential in vector mathematics. A vector's magnitude, often called its length or norm, is a measure of how long the vector is. To calculate the magnitude of a vector \( \mathbf{v} = [x, y, z] \), we apply the formula:

• \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \)

This equation essentially extends Pythagoras' theorem into three dimensions. In our example, the vector \( \mathbf{v} = [3, -2, 1] \) has a magnitude calculated as:

• \( \|\mathbf{v}\| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14} \)

This magnitude tells us how "big" or "intense" the vector is in space.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (a real number), scaling its magnitude without changing its direction unless the scalar is negative. If \( \mathbf{v} = [x, y, z] \) and \( k \) is a scalar, then:

• \( k\mathbf{v} = [kx, ky, kz] \)

This operation scales the vector without affecting its direction. If \( k > 0 \), the vector's direction remains unchanged. If \( k < 0 \), the direction is reversed. In our problem, the vector \( \mathbf{u} \) is a scalar multiple of \( \mathbf{v} \), satisfying \( \|\mathbf{u}\| = 3\|\mathbf{v}\| \). Through calculation, \( k = 3 \) or \( k = -3 \), leading to vectors \([9, -6, 3]\) and \([-9, 6, -3]\).

Parallel Vectors

Two vectors are parallel if they have the same direction or are exact opposites. This means one vector can be expressed as a scalar multiple of the other. In formal terms:

• If \( \mathbf{u} = k\mathbf{v} \), then \( \mathbf{u} \) and \( \mathbf{v} \) are parallel.

Parallel vectors do not intersect and maintain a constant direction relation. In the example, since \( \mathbf{u} = k\mathbf{v} \) with \( k = 3 \) or \( k = -3 \), the vector \( \mathbf{u} \) is indeed parallel to \( \mathbf{v} \). This confirms that the direction, magnitude, and opposition (if negative) are preserved. Hence, parallel vectors reveal the idea of scalar transformation maintaining inherent directional properties.