شرح مقدار المتجه

Worked example

1. 1

   $\mathbf{a}=\left(\begin{array}{c}-5\\ 12\end{array}\right)$ — The top number is 5 across (to the left) and the bottom is 12 up. They are the two short sides of a right-angled triangle.

2. 2

   $(-5{)}^2=25$ — Square the top number. Keep the brackets, because a negative times a negative is positive.

3. 3

   $12^2=144$ — Square the bottom number.

4. 4

   $25+144=169$ — Add the two squares. This is the long side squared.

5. 5

   $|\mathbf{a}|=\sqrt{169}$ — Take the square root to get back to a length.

6. 6

   $|\mathbf{a}|=13$ — A length is positive, so there is only one answer.

Worked example

1. 1

   $\overrightarrow{AB}=\left(\begin{array}{c}7\\ -2\end{array}\right),\overrightarrow{CB}=\left(\begin{array}{c}-1\\ 4\end{array}\right)$ — You are given AB and CB but asked for the length of AC. Find the vector AC first.

2. 2

   $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$ — Walk from A to B, then from B to C.

3. 3

   $\overrightarrow{BC}=-\overrightarrow{CB}=\left(\begin{array}{c}1\\ -4\end{array}\right)$ — You are given CB, which goes the other way. Reverse it by changing both signs.

4. 4

   $\overrightarrow{AC}=\left(\begin{array}{c}7\\ -2\end{array}\right)+\left(\begin{array}{c}1\\ -4\end{array}\right)$ — Put in the two columns.

5. 5

   $\overrightarrow{AC}=\left(\begin{array}{c}7+1\\ -2+(-4)\end{array}\right)=\left(\begin{array}{c}8\\ -6\end{array}\right)$ — Add the tops, and add the bottoms.

6. 6

   $|\overrightarrow{AC}{|}^2=8^2+(-6{)}^2=64+36$ — Now use Pythagoras on the new vector. Square each number, and the negative becomes positive.

7. 7

   $|\overrightarrow{AC}|=\sqrt{100}=10$ — Add to get 100, then take the square root.

Worked example

1. 1

   $P(-1,5),Q(4,-3)$ — The distance between two points is the length of the vector that joins them.

2. 2

   $\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}$ — Use end minus start to find the vector from P to Q.

3. 3

   $\overrightarrow{PQ}=\left(\begin{array}{c}4\\ -3\end{array}\right)-\left(\begin{array}{c}-1\\ 5\end{array}\right)$ — Put in the two columns, with brackets round the negative number.

4. 4

   $4-(-1)=5,-3-5=-8$ — Subtract the tops, then the bottoms. Taking away −1 adds 1.

5. 5

   $|\overrightarrow{PQ}{|}^2=5^2+(-8{)}^2=25+64$ — Square each number. The −8 squared is positive.

6. 6

   $|\overrightarrow{PQ}|=\sqrt{89}=9.43$ — Add to get 89, take the square root, and round to 3 significant figures.

Worked example

1. 1

   $\overrightarrow{AB}=\left(\begin{array}{c}-8\\ -15\end{array}\right),\overrightarrow{EF}=-2\overrightarrow{AB}$ — You want |EF|. Find the length of AB first, because it is quick.

2. 2

   $|\overrightarrow{AB}{|}^2=(-8{)}^2+(-15{)}^2=64+225$ — Square each number. Both negatives become positive.

3. 3

   $|\overrightarrow{AB}|=\sqrt{289}=17$ — Add to get 289, then take the square root.

4. 4

   $|\overrightarrow{EF}|=|-2|\times 17$ — Use the rule for a multiple.

5. 5

   $|\overrightarrow{EF}|=2\times 17=34$ — A length is never negative, so the −2 becomes 2.

6. 6

   $\left(\begin{array}{c}16\\ 30\end{array}\right):16^2+30^2=1156=34^2$ — Check it: −2AB is this column, and its length is 34.

Worked example

1. 1

   $\overrightarrow{DA}=\left(\begin{array}{c}-1\\ 2\end{array}\right),\overrightarrow{DB}=\left(\begin{array}{c}3\\ 4\end{array}\right),\overrightarrow{DC}=3\overrightarrow{AB}$ — These are the facts. AB is not given, so build it from DA and DB.

2. 2

   $\overrightarrow{AB}=\overrightarrow{AD}+\overrightarrow{DB}$ — Walk from A to D, then from D to B.

3. 3

   $\overrightarrow{AD}=-\overrightarrow{DA}=\left(\begin{array}{c}1\\ -2\end{array}\right)$ — DA goes the other way, so change both signs.

4. 4

   $\overrightarrow{AB}=\left(\begin{array}{c}1\\ -2\end{array}\right)+\left(\begin{array}{c}3\\ 4\end{array}\right)=\left(\begin{array}{c}4\\ 2\end{array}\right)$ — Add the tops, then the bottoms.

5. 5

   $\overrightarrow{DC}=3\overrightarrow{AB}=3\left(\begin{array}{c}4\\ 2\end{array}\right)$ — The question says DC is 3 times AB.

6. 6

   $\overrightarrow{DC}=\left(\begin{array}{c}12\\ 6\end{array}\right)$ — Multiply each number by 3.

7. 7

   $\overrightarrow{BC}=\overrightarrow{BD}+\overrightarrow{DC}$ — You want BC. Walk from B to D, then from D to C.

8. 8

   $\overrightarrow{BD}=-\overrightarrow{DB}=\left(\begin{array}{c}-3\\ -4\end{array}\right)$ — DB goes the other way, so change both signs.

9. 9

   $\overrightarrow{BC}=\left(\begin{array}{c}-3\\ -4\end{array}\right)+\left(\begin{array}{c}12\\ 6\end{array}\right)=\left(\begin{array}{c}9\\ 2\end{array}\right)$ — Add the tops, then the bottoms.

10. 10

    $|\overrightarrow{BC}|=\sqrt{9^2+2^2}=\sqrt{81+4}$ — Square each number and add.

11. 11

    $|\overrightarrow{BC}|=\sqrt{85}$ — The question says exact, so leave the root. 85 = 5 × 17 has no square number in it, so this cannot be simplified.

Worked example

1. 1

   $\mathbf{v}=\left(\begin{array}{c}2\\ -4\end{array}\right)$ — Find |v| exactly.

2. 2

   $|\mathbf{v}|=\sqrt{2^2+(-4{)}^2}=\sqrt{4+16}$ — Square each number and add. The negative becomes positive.

3. 3

   $|\mathbf{v}|=\sqrt{20}$ — Add, and stop before the root.

4. 4

   $\sqrt{20}=\sqrt{4\times 5}$ — Write 20 as a square number times another number. 4 is the biggest square that goes into 20.

5. 5

   $\sqrt{4\times 5}=\sqrt{4}\times \sqrt{5}$ — The root of a product is the product of the roots.

6. 6

   $|\mathbf{v}|=2\sqrt{5}$ — √4 is 2. This is the exact answer, with nothing left to take out of the 5.

Worked example

1. 1

   $\left|\left(\begin{array}{c}k\\ 8\end{array}\right)\right|=10$ — The length is given and the top number is missing.

2. 2

   $k^2+8^2=10^2$ — Square each number on the left and the length on the right. Squaring both sides gets rid of the root.

3. 3

   $k^2+64=100$ — 8² = 64 and 10² = 100.

4. 4

   $k^2=36$ — Take 64 from both sides.

5. 5

   $k=6\text{or}-6$ — Take the square root. There are two answers, because 6² and (−6)² are both 36.

Worked example

1. 1

   $\left|\left(\begin{array}{c}5\\ \sqrt{y}\end{array}\right)\right|=7$ — Now the missing number is inside a square root.

2. 2

   $5^2+(\sqrt{y}{)}^2=7^2$ — Square each number and the length. Squaring a square root gives the number back, so (√y)² is just y.

3. 3

   $25+y=49$ — 5² = 25 and 7² = 49.

4. 4

   $y=24$ — Take 25 from both sides. Check: 25 + 24 = 49 and √49 = 7.