Multiplying a Vector by a Scalar

Worked example

1. 1

   $\mathbf{a}=\left(\begin{array}{c}2\\ -1\end{array}\right),\mathbf{b}=\left(\begin{array}{c}-1\\ 4\end{array}\right)$ — We want 2a − 3b. Scale each vector first, then subtract.

2. 2

   $2\mathbf{a}=\left(\begin{array}{c}2\times 2\\ 2\times (-1)\end{array}\right)=\left(\begin{array}{c}4\\ -2\end{array}\right)$ — Multiply both numbers by 2.

3. 3

   $3\mathbf{b}=\left(\begin{array}{c}3\times (-1)\\ 3\times 4\end{array}\right)=\left(\begin{array}{c}-3\\ 12\end{array}\right)$ — Multiply both numbers by 3. Do all the scaling before any adding or subtracting.

4. 4

   $2\mathbf{a}-3\mathbf{b}=\left(\begin{array}{c}4-(-3)\\ -2-12\end{array}\right)$ — Subtract top from top and bottom from bottom.

5. 5

   $4-(-3)=4+3=7$ — Top row: taking away a negative adds.

6. 6

   $-2-12=-14$ — Bottom row: take 12 away from −2.

7. 7

   $2\mathbf{a}-3\mathbf{b}=\left(\begin{array}{c}7\\ -14\end{array}\right)$ — Put both answers in one column.

Worked example

1. 1

   $H(2,-3),\overrightarrow{HJ}=\left(\begin{array}{c}4\\ -5\end{array}\right),\overrightarrow{HK}=3\overrightarrow{HJ}$ — We want the point K.

2. 2

   $\overrightarrow{HK}=\left(\begin{array}{c}3\times 4\\ 3\times (-5)\end{array}\right)=\left(\begin{array}{c}12\\ -15\end{array}\right)$ — HK is three times HJ, so multiply both numbers by 3.

3. 3

   $2+12=14$ — HK starts at H, so add its top number to the x-coordinate of H.

4. 4

   $-3+(-15)=-18$ — Add its bottom number to the y-coordinate of H.

5. 5

   $K=(14,-18)$ — Write the answer as a point.

Worked example

1. 1

   $A(1,2),B(4,-1),\overrightarrow{AC}=2\overrightarrow{BA}$ — We need BA first, which goes from B to A.

2. 2

   $\overrightarrow{BA}=\left(\begin{array}{c}1-4\\ 2-(-1)\end{array}\right)=\left(\begin{array}{c}-3\\ 3\end{array}\right)$ — End minus start: A is the end and B is the start.

3. 3

   $\overrightarrow{AC}=2\left(\begin{array}{c}-3\\ 3\end{array}\right)=\left(\begin{array}{c}-6\\ 6\end{array}\right)$ — Double both numbers.

4. 4

   $1+(-6)=-5$ — AC starts at A, so add its top number to the x-coordinate of A.

5. 5

   $2+6=8$ — Add its bottom number to the y-coordinate of A.

6. 6

   $C=(-5,8)$ — Write the answer as a point.

Worked example

1. 1

   $\left(\begin{array}{c}6\\ -9\end{array}\right)\text{and}\left(\begin{array}{c}2\\ -3\end{array}\right)$ — Are they parallel? Look for one number that turns the second into the first.

2. 2

   $6÷2=3$ — Divide the top of the first by the top of the second.

3. 3

   $-9÷(-3)=3$ — Do the same with the bottoms. A negative divided by a negative is positive.

4. 4

   $\left(\begin{array}{c}6\\ -9\end{array}\right)=3\left(\begin{array}{c}2\\ -3\end{array}\right)$ — Both rows give the same number 3, so the first is 3 times the second. They are parallel.

Worked example

1. 1

   $\left(\begin{array}{c}-4\\ p\end{array}\right)\text{parallel to}\left(\begin{array}{c}2\\ -3\end{array}\right)$ — Find p. The first vector must be a number times the second.

2. 2

   $-4÷2=-2$ — Use the tops, where both numbers are known, to find the scalar.

3. 3

   $\left(\begin{array}{c}-4\\ p\end{array}\right)=-2\left(\begin{array}{c}2\\ -3\end{array}\right)$ — The scalar is −2, so the bottom number must be scaled by −2 too.

4. 4

   $p=-2\times (-3)=6$ — A negative times a negative is positive.

Worked example

1. 1

   $6\mathbf{a}-9\mathbf{b}$ — Look for a number that divides both 6 and 9. The biggest is 3.

2. 2

   $=3(2\mathbf{a}-3\mathbf{b})$ — Take out 3: 6 ÷ 3 = 2 and 9 ÷ 3 = 3, and the minus stays.

3. 3

   $-4\mathbf{a}+6\mathbf{b}$ — Now one that starts with a minus. We want the bracket to match 2a − 3b.

4. 4

   $=-2(2\mathbf{a}-3\mathbf{b})$ — Take out −2: −4 ÷ (−2) = 2 and 6 ÷ (−2) = −3.