Scalar Multiplication of Vectors

How a Real Multiplier Changes a Vector

The result is a new vector whose magnitude is scaled by the scalar's absolute value. For a nonzero original vector, the sign determines whether the direction stays the same or reverses.

If $k$ is a real number (scalar) and $\vec{v}$ is a vector, then the scalar multiplication of a vector is denoted as $k \cdot \vec{v}$.

Properties of Scalar Multiplication of Vectors

Scalar multiplication changes a vector according to the following rules:

1. If $k > 0$ and the original vector is nonzero, the resulting vector has the same direction.
2. If $k < 0$ and the original vector is nonzero, the resulting vector points in the opposite direction.
3. If $k = 0$, then the resulting vector is a zero vector.
4. The resulting magnitude is the original magnitude multiplied by the factor $|k|$.

Representation of Scalar Multiplication of Vectors

Multiplying a vector by a number changes its length by that factor and may also reverse it. The result stays a vector on the same line, so the length and the direction can be checked separately.

Magnitude and Direction of a Scaled Vector

Geometrically, scalar multiplication changes a vector's magnitude by the factor $|k|$. For a nonzero vector, its direction depends on the sign of $k$:

• If $k > 0$, the direction of the vector remains unchanged
• If $k < 0$, the direction of the vector is opposite to the original vector

In both cases, and also when $k=0$, the magnitude satisfies:

Scaling a Vector by Components

If $\vec{v} = (v_1, v_2, v_3)$ is a vector in $3$-dimensional space, then:

In unit vector notation:

Multiplying Vectors by Positive and Negative Scalars

A factor greater than one stretches the vector, a factor between zero and one shrinks it, and a negative factor also reverses the direction. So multiplying by a negative scalar always changes the direction, while multiplying by a positive scalar never does.

Multiplying by a Positive Scalar

Given the vector $\vec{a} = 2\vec{i} + 3\vec{j} + 5\vec{k}$. Determine the result of multiplication $2\vec{a}$.

Solution:

Multiplying by a Negative Scalar

Given the vector $\vec{v} = (4, -2, 3)$. Determine the result of $-3\vec{v}$.

Solution:

The resulting vector points in the opposite direction because the scalar is negative.

Applications of Scalar Multiplication of Vectors

Scalar multiplication of vectors has many applications in physics and mathematics, such as:

1. Force and Acceleration: If an object with mass $m$ experiences acceleration $\vec{a}$, then the force acting on the object is $\vec{F} = m\vec{a}$.

2. Velocity: If an object moves with constant velocity $\vec{v}$ for a time $t$, then its displacement is $\vec{s} = t\vec{v}$.

3. Scaling in Computer Graphics: Multiplying every position vector by $k$ scales an object about the coordinate origin. To scale about another center $\vec{c}$, each point $\vec{p}$ must be translated, scaled, and translated back according to $\vec{p}'=\vec{c}+k(\vec{p}-\vec{c})$.

Practice Problems

Each problem multiplies a vector by one or two scalars. Multiply every component by the same factor, then combine the results where two scalars appear.

1. Given the vector $\vec{a} = 3\vec{i} - 4\vec{j} + 2\vec{k}$. Determine the result of $-2\vec{a}$.

2. Vectors $\vec{u} = (2, -5, 1)$ and $\vec{v} = (-4, 3, 6)$. Determine the vector $2\vec{u} - 3\vec{v}$.

3. Given $|\overrightarrow{BR}| = 3.4 \text{ cm}$. If $\overrightarrow{BU} = 0.65 \cdot \overrightarrow{BR}$ and $\overrightarrow{UR} = 0.35 \cdot \overrightarrow{BR}$, prove that all three vectors have the same direction.

4. Vector $\vec{p}$ has a length of $5 \text{ units}$ and vector $\vec{q} = 3\vec{p}$. Determine the length of vector $\vec{q}$.

5. Given points $A(2, 3, -1)$, $B(5, -2, 4)$, and $C$ lies on the line passing through $A$ and $B$ such that $\overrightarrow{AC} = 2\overrightarrow{AB}$. Determine the coordinates of point $C$.

Answer Key

Each answer keeps the full computation and the final value, so you can compare your own attempt line by line. Read the working first and check your numbers afterwards.

Each answer multiplies component by component and then recombines the terms. Compare your result with the value at each coordinate.

First Problem

Given the vector $\vec{a} = 3\vec{i} - 4\vec{j} + 2\vec{k}$. Determine the result of $-2\vec{a}$.

Solution:

The result of $-2\vec{a}$ is $-6\vec{i} + 8\vec{j} - 4\vec{k}$.

Second Problem

Vectors $\vec{u} = (2, -5, 1)$ and $\vec{v} = (-4, 3, 6)$. Determine the vector $2\vec{u} - 3\vec{v}$.

Solution:

The vector $2\vec{u} - 3\vec{v}$ is $(16, -19, -16)$ or $16\vec{i} - 19\vec{j} - 16\vec{k}$.

Third Problem

Given $|\overrightarrow{BR}| = 3.4 \text{ cm}$. If $\overrightarrow{BU} = 0.65 \cdot \overrightarrow{BR}$ and $\overrightarrow{UR} = 0.35 \cdot \overrightarrow{BR}$, prove that all three vectors have the same direction.

Solution: To prove that all three vectors have the same direction, we need to show that they are positive scalar multiples of the same vector.

We know:

• $\overrightarrow{BU} = 0.65 \cdot \overrightarrow{BR}$
• $\overrightarrow{UR} = 0.35 \cdot \overrightarrow{BR}$

Check if $\overrightarrow{BU} + \overrightarrow{UR} = \overrightarrow{BR}$:

The equality $\overrightarrow{BU} + \overrightarrow{UR} = \overrightarrow{BR}$ is the vector addition law for the collinear points $B$, $U$, and $R$.

Since $\overrightarrow{BU} = 0.65 \cdot \overrightarrow{BR}$ and $\overrightarrow{UR} = 0.35 \cdot \overrightarrow{BR}$, where the scalar factors are positive ($0.65$ and $0.35$), all three vectors have the same direction. Positive scalar factors mean that these vectors point in the same direction as the reference vector $\overrightarrow{BR}$.

Fourth Problem

Vector $\vec{p}$ has a length of $5 \text{ units}$ and vector $\vec{q} = 3\vec{p}$. Determine the length of vector $\vec{q}$.

Solution: Given $|\vec{p}| = 5 \text{ units}$ and $\vec{q} = 3\vec{p}$.

To determine the length of vector $\vec{q}$, we use the property of scalar multiplication:

The length of vector $\vec{q}$ is $15 \text{ units}$.

Fifth Problem

Given points $A(2, 3, -1)$, $B(5, -2, 4)$, and $C$ lies on the line passing through $A$ and $B$ such that $\overrightarrow{AC} = 2\overrightarrow{AB}$. Determine the coordinates of point $C$.

Solution: First, we determine the vector $\overrightarrow{AB}$:

Then, we use the relationship $\overrightarrow{AC} = 2\overrightarrow{AB}$ to determine the vector $\overrightarrow{AC}$:

Next, we determine the coordinates of point $C$:

The coordinates of point $C$ are $C(8, -7, 9)$.