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Adding vectors Algebraic expressionHere you will learn about vector multiplication, including scalar multiplication of a vector (multiplication of a vector by a number).
Students will first learn about vector multiplication as part of number and quantity in high school.
Vector multiplication is when you multiply a vector by a number called a scalar. The scalar has magnitude only, whereas a vector has magnitude and direction.
To do this, you multiply each component of the vector by the scalar.
For example,
Vector \textbf{a} is \textbf{a} = \begin{pmatrix} \; 2 \; \\ \; 1 \; \end{pmatrix} .
So multiplying the vector \textbf{a} by 3 would be
3\textbf{a} = 3 \begin{pmatrix} \; 2 \; \\ \; 1 \; \end{pmatrix} = \begin{pmatrix} \; 3 \times 2 \; \\ \; 3 \times 1 \; \end{pmatrix} = \begin{pmatrix} \; 6 \; \\ \; 3 \; \end{pmatrix}The image below shows a visual representation of the vectors \textbf{a} and 3\textbf{a}.
This works because multiplication is repeated addition. Another way of thinking about multiplying the vector \textbf{a} by 3 would be
3\textbf{a}=\textbf{a}+\textbf{a}+\textbf{a}= \begin{pmatrix} \; 2 \; \\ \; 1 \; \end{pmatrix} + \begin{pmatrix} \; 2 \; \\ \; 1 \; \end{pmatrix} + \begin{pmatrix} \; 2 \; \\ \; 1 \; \end{pmatrix} = \begin{pmatrix} \; 6 \; \\ \; 3 \; \end{pmatrix}You can see 3 groups of the x component and 3 groups of the y component on the diagram.
When multiplied by a scalar number, the direction of the vector is the same as the original vector, but the magnitude of the vector (also called the absolute value of the vector) has changed.
In general,
a \begin{pmatrix} \; b \; \\ \; c \; \end{pmatrix} = \begin{pmatrix} \; b \times a \; \\ \; c \times a \; \end{pmatrix} = \begin{pmatrix} \; ab \; \\ \; ac \; \end{pmatrix}
for any value of a, \, b, and c.
How does this relate to high school math?
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DOWNLOAD FREEIn order to multiply a vector by a scalar:
Solve 2 \begin{pmatrix} \; 4 \; \\ \; 5 \; \end{pmatrix} .
Multiply the scalar number by the top number.
2\times4=82Multiply the \textbf{y} component by the scalar.
Multiply the scalar number by the bottom number.
2\times5=103Write the resultant vector.
Write the two answers as a column vector.
2 \begin{pmatrix} \; 4 \; \\ \; 5 \; \end{pmatrix} = \begin{pmatrix} \; 2 \times 4 \; \\ \; 2 \times 5 \; \end{pmatrix} = \begin{pmatrix} \; 8 \; \\ \; 10 \; \end{pmatrix}Solve 4 \begin{pmatrix} \; 8 \; \\ \; 3 \; \end{pmatrix} .
Multiply the \textbf{x} component by the scalar.
Multiply the scalar number by the top number.
4\times8=32
Multiply the \textbf{y} component by the scalar.
Multiply the scalar number by the bottom number.
4\times3=12
Write the resultant vector.
Write the two answers as a column vector.
4 \begin{pmatrix} \; 8 \; \\ \; 3 \; \end{pmatrix} = \begin{pmatrix} \; 4 \times 8 \; \\ \; 4 \times 3 \; \end{pmatrix} = \begin{pmatrix} \; 32 \; \\ \; 12 \; \end{pmatrix}
Solve 0.5 \begin{pmatrix} \; 10 \; \\ \; 8 \; \end{pmatrix} .
Multiply the \textbf{x} component by the scalar.
Multiply the scalar number by the top number.
0.5\times10=5
Multiply the \textbf{y} component by the scalar.
Multiply the scalar number by the bottom number.
0.5\times8=4
Write the resultant vector.
Write the two answers as a column vector.
0.5 \begin{pmatrix} \; 10 \; \\ \; 8 \; \end{pmatrix} = \begin{pmatrix} \; 0.5 \times 10 \; \\ \; 0.5 \times 8 \; \end{pmatrix} = \begin{pmatrix} \; 5 \; \\ \; 4 \; \end{pmatrix}
Solve \cfrac{1}{3} \begin{pmatrix} \; - \, 6 \; \\ \; 9 \; \end{pmatrix} .
Multiply the \textbf{x} component by the scalar.
Multiply the scalar number by the top number.
\cfrac{1}{3}\times - \, 6=- \, 2
Multiply the \textbf{y} component by the scalar.
Multiply the scalar number by the bottom number.
\cfrac{1}{3}\times9=3
Write the resultant vector.
Write the two answers as a column vector.
\cfrac{1}{3} \begin{pmatrix} \; - \, 6 \; \\ \; 9 \; \end{pmatrix} = \begin{pmatrix} \; \cfrac{1}{3} \times - \, 6 \; \\ \; \cfrac{1}{3} \times 9 \; \end{pmatrix} = \begin{pmatrix} \; - \, 2 \; \\ \; 3 \; \end{pmatrix}
Solve 4 \begin{pmatrix} \; - \, 5 \; \\ \; - \, 2 \; \end{pmatrix} .
Multiply the \textbf{x} component by the scalar.
Multiply the scalar number by the top number.
4\times - \, 5=- \, 20
Multiply the \textbf{y} component by the scalar.
Multiply the scalar number by the bottom number.
4\times- \, 2=- \, 8
Write the resultant vector.
Write the two answers as a column vector.
4 \begin{pmatrix} \; - \, 5 \; \\ \; - \, 2 \; \end{pmatrix} = \begin{pmatrix} \; 4 \times - \, 5 \; \\ \; 4 \times - \, 2 \; \end{pmatrix} = \begin{pmatrix} \; - \, 20 \; \\ \; - \, 8 \; \end{pmatrix}Solve - \, 3 \begin{pmatrix} \; - \, 2 \; \\ \; 5 \; \end{pmatrix} .
Multiply the \textbf{x} component by the scalar.
Multiply the scalar number by the top number.
- \, 3\times- \, 2=6
Multiply the \textbf{y} component by the scalar.
Multiply the scalar number by the bottom number.
- \, 3\times5=- \, 15
Write the resultant vector.
Write the two answers as a column vector.
- \, 3 \begin{pmatrix} \; - \, 2 \; \\ \; 5 \; \end{pmatrix} = \begin{pmatrix} \; - \, 3 \times - \, 2 \; \\ \; - \, 3 \times 5 \; \end{pmatrix} = \begin{pmatrix} \; 6 \; \\ \; - \, 15 \; \end{pmatrix}
Solve 3 \begin{pmatrix} \; x \; \\ \; - \, y \; \end{pmatrix} .
Multiply the \textbf{x} component by the scalar.
Multiply the scalar number by the top number.
3\times{x}=3x
Multiply the \textbf{y} component by the scalar.
Multiply the scalar number by the bottom number.
3\times{- \, y}=- \, 3y
Write the resultant vector.
Write the two answers as a column vector.
3 \begin{pmatrix} \; x \; \\ \; - \, y \; \end{pmatrix} = \begin{pmatrix} \; x \times 3 \; \\ \; - \, y \times 3 \; \end{pmatrix} = \begin{pmatrix} \; 3x \; \\ \; - \, 3y \; \end{pmatrix}
1. Find 3\textbf{a} when \textbf{a} = \begin{pmatrix} \; 2 \; \\ \; 4 \; \end{pmatrix}.
2. Find 5\textbf{b} when vector \textbf{b} is \textbf{b} = \begin{pmatrix} \; 1 \; \\ \; 3 \; \end{pmatrix}.
3. Find 0.1\textbf{c} when \textbf{c} = \begin{pmatrix} \; 20 \; \\ \; 30 \; \end{pmatrix}.
4. Find \cfrac{2}{5} \, \textbf{a} when \textbf{a} = \begin{pmatrix} \; – \, 10 \; \\ \; 15 \; \end{pmatrix}.
5. Find 6\textbf{d} when \textbf{d} = \begin{pmatrix} \; – \, 2 \; \\ \; – \, 3 \; \end{pmatrix}.
6. Find – \, 3\textbf{f} when \textbf{f} = \begin{pmatrix} \; – \, 5 \; \\ \; – \, 4 \; \end{pmatrix}.
7. Find 2\textbf{g} when \textbf{g} = \begin{pmatrix} \; 2b \; \\ \; 7c \; \end{pmatrix}.
Vector multiplication by a scalar involves multiplying each component of a vector by the scalar value. This operation changes the magnitude of the vector, but the direction remains the same if the scalar is positive. If the scalar is negative, the direction of the vector is reversed.
No, multiplying two vectors together requires different operations, such as the dot product or cross product of two vectors, depending on the context. Vector-scalar multiplication is simpler because it only involves scaling the vector by a single number (scalar).
Vector multiplication by a scalar is not directly related to trigonometry because it doesn’t involve angles. However, vector multiplication is related to trigonometry through operations like the dot product and cross product, which both involve angles between vectors.
The dot product uses the cosine of the angle to measure alignment, while the cross product uses the sine of the angle to determine the perpendicular vector’s magnitude.
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