Scalar Multiplication Notes

Published on: September 29, 2026

This Scalar Multiplication Calculator multiplies a vector by a single number, and shows the working line by line. Every component is scaled by the same factor, which stretches or shrinks the vector without turning it. A negative scalar also flips it to point the opposite way. Following the same pattern every time makes it clear that direction only ever changes when the scalar is negative.

Step-by-step method

Set up the problem.
Write the scalar multiplication rule.
Substitute values into the rule.
Multiply each component and simplify.

See Example 1 Hide Example 1

Problem

\(k = 3,\; \vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle\)

Approach

Step-by-step method

Set up the problem.
Write the scalar multiplication rule.
Substitute values into the rule.
Multiply each component and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The scalar is \(k = 3\) and the vector is \(\vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle\).

\(k = 3,\; \vec{A} = \left\langle 1,\; 2,\; 3 \right\rangle\)

Step 2

Step 2 - Write the scalar multiplication rule.

In this problem: Multiply every component of the vector by the scalar.

\(k\vec{A} = \left\langle k a_{1},\; k a_{2},\; k a_{3} \right\rangle\)

Step 3

Step 3 - Substitute values into the rule.

In this problem: Substitute the scalar and each component.

\(k\vec{A} = \left\langle 3 \cdot 1,\; 3 \cdot 2,\; 3 \cdot 3 \right\rangle\)

Step 4

Step 4 - Multiply each component and simplify.

In this problem: The resulting vector is \(\left\langle 3,\; 6,\; 9 \right\rangle\).

\(k\vec{A} = \left\langle 3,\; 6,\; 9 \right\rangle\)

Final Answer

\(k\vec{A} = \left\langle 3,\; 6,\; 9 \right\rangle\)

See Example 2 Hide Example 2

Problem

\(k = -2,\; \vec{A} = \left\langle \frac{1}{2},\; 0,\; 4 \right\rangle\)

Approach

Step-by-step method

Set up the problem.
Write the scalar multiplication rule.
Substitute values into the rule.
Multiply each component and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The scalar is \(k = -2\) and the vector is \(\vec{A} = \left\langle \frac{1}{2},\; 0,\; 4 \right\rangle\).

\(k = -2,\; \vec{A} = \left\langle \frac{1}{2},\; 0,\; 4 \right\rangle\)

Step 2

Step 2 - Write the scalar multiplication rule.

In this problem: Multiply every component of the vector by the scalar.

\(k\vec{A} = \left\langle k a_{1},\; k a_{2},\; k a_{3} \right\rangle\)

Step 3

Step 3 - Substitute values into the rule.

In this problem: Substitute the scalar and each component.

\(k\vec{A} = \left\langle \left(-2\right) \cdot \frac{1}{2},\; \left(-2\right) \cdot 0,\; \left(-2\right) \cdot 4 \right\rangle\)

Step 4

Step 4 - Multiply each component and simplify.

In this problem: The resulting vector is \(\left\langle -1,\; 0,\; -8 \right\rangle\).

\(k\vec{A} = \left\langle -1,\; 0,\; -8 \right\rangle\)

Final Answer

\(k\vec{A} = \left\langle -1,\; 0,\; -8 \right\rangle\)

See Example 3 Hide Example 3

Problem

\(k = 4,\; \vec{A} = \left\langle 2,\; -1,\; 3 \right\rangle\)

Approach

Step-by-step method

Set up the problem.
Write the scalar multiplication rule.
Substitute values into the rule.
Multiply each component and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The scalar is \(k = 4\) and the vector is \(\vec{A} = \left\langle 2,\; -1,\; 3 \right\rangle\).

\(k = 4,\; \vec{A} = \left\langle 2,\; -1,\; 3 \right\rangle\)

Step 2

Step 2 - Write the scalar multiplication rule.

In this problem: Multiply every component of the vector by the scalar.

\(k\vec{A} = \left\langle k a_{1},\; k a_{2},\; k a_{3} \right\rangle\)

Step 3

Step 3 - Substitute values into the rule.

In this problem: Substitute the scalar and each component.

\(k\vec{A} = \left\langle 4 \cdot 2,\; 4 \cdot \left(-1\right),\; 4 \cdot 3 \right\rangle\)

Step 4

Step 4 - Multiply each component and simplify.

In this problem: The resulting vector is \(\left\langle 8,\; -4,\; 12 \right\rangle\).

\(k\vec{A} = \left\langle 8,\; -4,\; 12 \right\rangle\)

Final Answer

\(k\vec{A} = \left\langle 8,\; -4,\; 12 \right\rangle\)

See Example 4 Hide Example 4

Problem

\(k = -3,\; \vec{A} = \left\langle 1,\; 5,\; -2 \right\rangle\)

Approach

Step-by-step method

Set up the problem.
Write the scalar multiplication rule.
Substitute values into the rule.
Multiply each component and simplify.

Step 1

Step 1 - Set up the problem.

In this problem: The scalar is \(k = -3\) and the vector is \(\vec{A} = \left\langle 1,\; 5,\; -2 \right\rangle\).

\(k = -3,\; \vec{A} = \left\langle 1,\; 5,\; -2 \right\rangle\)

Step 2

Step 2 - Write the scalar multiplication rule.

In this problem: Multiply every component of the vector by the scalar.

\(k\vec{A} = \left\langle k a_{1},\; k a_{2},\; k a_{3} \right\rangle\)

Step 3

Step 3 - Substitute values into the rule.

In this problem: Substitute the scalar and each component.

\(k\vec{A} = \left\langle \left(-3\right) \cdot 1,\; \left(-3\right) \cdot 5,\; \left(-3\right) \cdot \left(-2\right) \right\rangle\)

Step 4

Step 4 - Multiply each component and simplify.

In this problem: The resulting vector is \(\left\langle -3,\; -15,\; 6 \right\rangle\).

\(k\vec{A} = \left\langle -3,\; -15,\; 6 \right\rangle\)

Final Answer

\(k\vec{A} = \left\langle -3,\; -15,\; 6 \right\rangle\)