Vector Arithmetic (SQA National 5 Maths): Revision Note

Adding & subtracting vector components

How do I add and subtract vectors in component form?

• Adding and subtracting vectors is done by looking at the components of the vectors separately

• To add vectors in component form

  • Add the x components together

  • Add the y components together

  • Add the z components together (only if the vector is three-dimensional!)

    • $\left(\begin{array}{c}5\\ 2\end{array}\right)+\left(\begin{array}{c}3\\ -1\end{array}\right)=\left(\begin{array}{c}5+3\\ 2+\left(-1\right)\end{array}\right)=\left(\begin{array}{c}8\\ 1\end{array}\right)$

    • $\left(\begin{array}{c}2\\ -3\\ 5\end{array}\right)+\left(\begin{array}{c}1\\ 1\\ -4\end{array}\right)=\left(\begin{array}{c}2+1\\ -3+1\\ 5+\left(-4\right)\end{array}\right)=\left(\begin{array}{c}3\\ -2\\ 1\end{array}\right)$

• To subtract vectors in component form

  • Subtract the second x component from the first

  • Subtract the second y component from the first

  • Subtract the second z component from the first (only if the vector is three-dimensional!)

    • $\left(\begin{array}{c}5\\ 2\end{array}\right)-\left(\begin{array}{c}3\\ -1\end{array}\right)=\left(\begin{array}{c}5-3\\ 2-\left(-1\right)\end{array}\right)=\left(\begin{array}{c}2\\ 3\end{array}\right)$

    • $\left(\begin{array}{c}2\\ -3\\ 5\end{array}\right)-\left(\begin{array}{c}1\\ 1\\ -4\end{array}\right)=\left(\begin{array}{c}2-1\\ -3-1\\ 5-\left(-4\right)\end{array}\right)=\left(\begin{array}{c}1\\ -4\\ 9\end{array}\right)$

Multiplying a vector by a scalar

How do I multiply a vector by a scalar?

• A scalar is a regular number not a vector

  • It does not have a direction

• To multiply a column vector by a scalar

  • Multiply the x component by the scalar

  • Multiply the y component by the scalar

  • Multiply the z component by the scalar (only if the vector is three-dimensional!)

    • $3\left(\begin{array}{c}2\\ -1\end{array}\right)=\left(\begin{array}{c}3\times 2\\ 3\times \left(-1\right)\end{array}\right)=\left(\begin{array}{c}6\\ -3\end{array}\right)$

    • $4\left(\begin{array}{c}3\\ 2\\ -5\end{array}\right)=\left(\begin{array}{c}4\times 3\\ 4\times 2\\ 4\times \left(-5\right)\end{array}\right)=\left(\begin{array}{c}12\\ 8\\ -20\end{array}\right)$

How do I write an expression as a single vector in component form?

• You need to follow the order of operations

  • 2D: $2\left(\begin{array}{c}5\\ 2\end{array}\right)+5\left(\begin{array}{c}3\\ -1\end{array}\right)$

  • 3D: $3\left(\begin{array}{c}-1\\ 5\\ 2\end{array}\right)-7\left(\begin{array}{c}2\\ -2\\ 3\end{array}\right)$

• STEP 1
  Multiply each vector by the scalar in front of it

  • $\left(\begin{array}{c}2\times 5\\ 2\times 2\end{array}\right)+\left(\begin{array}{c}5\times 3\\ 5\times \left(-1\right)\end{array}\right)=\left(\begin{array}{c}10\\ 4\end{array}\right)+\left(\begin{array}{c}15\\ -5\end{array}\right)$

  • $\left(\begin{array}{c}3\times -1\\ 3\times 5\\ 3\times 2\end{array}\right)-\left(\begin{array}{c}7\times 2\\ 7\times -2\\ 7\times 3\end{array}\right)=\left(\begin{array}{c}-3\\ 15\\ 6\end{array}\right)-\left(\begin{array}{c}14\\ -14\\ 21\end{array}\right)$

• STEP 2
  Add or subtract the new column vectors

  • $\left(\begin{array}{c}10+15\\ 4+\left(-5\right)\end{array}\right)=\left(\begin{array}{c}25\\ -1\end{array}\right)$

  • $\left(\begin{array}{c}-3-14\\ 15-\left(-14\right)\\ 6-21\end{array}\right)=\left(\begin{array}{c}-17\\ 29\\ -15\end{array}\right)$

• The single vector at the end is known as a resultant vector

  • $\left(\begin{array}{c}25\\ -1\end{array}\right)$ is the resultant vector for $2\left(\begin{array}{c}5\\ 2\end{array}\right)+5\left(\begin{array}{c}3\\ -1\end{array}\right)$

  • $\left(\begin{array}{c}-17\\ 29\\ -15\end{array}\right)$ is the resultant vector for $3\left(\begin{array}{c}-1\\ 5\\ 2\end{array}\right)-7\left(\begin{array}{c}2\\ -2\\ 3\end{array}\right)$