vectors

vectors refers to the movement from one point to another. A vector quantity consists of direction and magnitude/length

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Deriving a vector from a graph

The graph below shows vector AB having direction and length. From the graph drawn to make up the components of a vector which is the x and y axis one must taken down the point where the vector lays on X and also on the Y axis

vector AB = [20, 30] or $\mathrm{AB}$ = $\left[\begin{array}{c}20\\ 30\end{array}\right]$

Points to note

A vector has both magnitude and direction but a scalar has magnitude only.

A vector may be represented by $\overrightarrow{\mathrm{OA}}$ or a like symbol.

The magnitude of a column vector a = $\left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\end{array}\right]$ is given by |a| = $\sqrt{x^2+y^2}$

Position Vectors

If the point P has coordinates (a, b), then the position vector of P, $\overrightarrow{\mathrm{OP}}$ , is written as $\overrightarrow{\mathrm{OP}}$ = $\left[\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right]$

Equal Vectors

Two vectors are equal when they have the same direction and magnitude.

Negative Vector

$\overrightarrow{\mathrm{BA}}$ is the negative of $\overrightarrow{\mathrm{AB}}$.

$\overrightarrow{\mathrm{BA}}$ is a vector having the same magnitude as $\overrightarrow{\mathrm{AB}}$ but having direction opposite to that of $\overrightarrow{\mathrm{AB}}$.

We can write $\overrightarrow{\mathrm{BA}}$ = $\overrightarrow{\mathrm{-\; AB}}$ and $\overrightarrow{\mathrm{AB}}$ = $\overrightarrow{\mathrm{-\; BA}}$.

Zero Vector

A vector whose magnitude is zero is called a zero vector and is denoted by 0.

Sum and Difference of Two Vectors

The sum of two vectors, a and b, can be determined by using the Triangle Law or Parallelogram Law of Vector Addition.

Triangle law of addition:

$\overrightarrow{\mathrm{AB}}$ + $\overrightarrow{\mathrm{BC}}$ = $\overrightarrow{\mathrm{AC}}$

Parallelogram law of addition:

$\overrightarrow{\mathrm{AB}}$ + $\overrightarrow{\mathrm{AD}}$ = $\overrightarrow{\mathrm{AB}}$ + $\overrightarrow{\mathrm{BC}}$ = $\overrightarrow{\mathrm{AC}}$

How to solve vectors step by step

Given that 𝒂 = $\left(\begin{array}{c}1\\ 2\end{array}\right)$,𝒃= $\left(\begin{array}{c}3\\ -1\end{array}\right)$ and 𝒄= $\left(\begin{array}{c}40\\ 0\end{array}\right)$, evaluate;

(i) a + b,

(ii) a – c,

(iii) c – b,

Solutions:

(i)

Addition of vectors

a + b

a = $\left(\begin{array}{c}1\\ 2\end{array}\right)$, b = $\left(\begin{array}{c}3\\ -1\end{array}\right)$

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

$\left(\begin{array}{c}\mathrm{1\; +\; 3}\\ \mathrm{2\; +\; -1}\end{array}\right)$

$\left(\begin{array}{c}\mathrm{1\; +\; 3}\\ \mathrm{2\; -\; 1}\end{array}\right)$

$\left(\begin{array}{c}4\\ 1\end{array}\right)$

Answer: a + b = $\left(\begin{array}{c}4\\ 1\end{array}\right)$

(ii)

Substraction of vectors

a - c

$\left(\begin{array}{c}1\\ 2\end{array}\right)$ - $\left(\begin{array}{c}40\\ 0\end{array}\right)$

$\left(\begin{array}{c}-39\\ 2\end{array}\right)$

Answer: $\left(\begin{array}{c}-39\\ 2\end{array}\right)$

(iii)

c – b

$\left(\begin{array}{c}40\\ 0\end{array}\right)$ - $\left(\begin{array}{c}3\\ -1\end{array}\right)$

$\left(\begin{array}{c}37\\ 1\end{array}\right)$

Answer: $\left(\begin{array}{c}37\\ 1\end{array}\right)$

How to solve vectors with examples

The diagram below shows a parallelogram where AB = p and BC = q, the point E on AD is such that AE = $\frac{1}{4}$AD

(i) Express in terms of p and/or q vectors

(a) AC

(b) AE

(c) BE

(ii) AC and BE intersect at F. Given that BF = kBE Express BF in terms of p, q and K.

(iii) Hence show that AF = (1 - k)p + 1/4kq

Solutions:

(i)

(a)

We first come up with the formula, what we want is to use to find AC

$\overrightarrow{\mathrm{AC}}$ = $\overrightarrow{\mathrm{AB}}$ + $\overrightarrow{\mathrm{BC}}$

substitute into the formula to find AC

$\overrightarrow{\mathrm{AC}}$ = p + q

Answer: $\overrightarrow{\mathrm{AC}}$ = p + q

(b)

Look at the first statement in the question, there is already a formula for AE which says, AE = $\frac{1}{4}$AD and having in mind that the diagram above is a parallelogram then AD = BC. Given tha BC = q

AE = $\frac{1}{4}$q

Answer: AE = $\frac{1}{4}$q

(c)

Formula for BE is :

$\overrightarrow{\mathrm{BE}}$ = $\overrightarrow{\mathrm{BA}}$ + $\overrightarrow{\mathrm{AE}}$

Note that: $\overrightarrow{\mathrm{BA}}$ is the negative vector of $\overrightarrow{\mathrm{AB}}$, hence $\overrightarrow{\mathrm{BA}}$ = -p

substitute in the formula

$\overrightarrow{\mathrm{BE}}$ = - p + $\frac{1}{4}$q

$\overrightarrow{\mathrm{BE}}$ = $\frac{1}{4}$q - p

Answer: $\overrightarrow{\mathrm{BE}}$ = $\frac{1}{4}$q - p

(ii)

Formula to solve BF = kBE and expressing it in terms of p, q and k

substitute in the formula given that $\overrightarrow{\mathrm{BE}}$ = $\frac{1}{4}$q - p

BF = k$\frac{1}{4}$q - p

BF = $\frac{1}{4}$kq - kp

Answer: $\overrightarrow{\mathrm{BF}}$ = $\frac{1}{4}$kq - kp

(iii)

We have to come up with a formula to prove that AF = (1 - k)p + 1/4kq

Formula

$\overrightarrow{\mathrm{AF}}$ = $\overrightarrow{\mathrm{AB}}$ + $\overrightarrow{\mathrm{BF}}$

We already have $\overrightarrow{\mathrm{AB}}$ = p and BF = $\frac{1}{4}$kq - kp, hence we can just substitute

$\overrightarrow{\mathrm{AF}}$ = p + $\frac{1}{4}$kq - kp

$\overrightarrow{\mathrm{AF}}$ = p - kp $\frac{1}{4}$kq

$\overrightarrow{\mathrm{AF}}$ = (1 - k)p $\frac{1}{4}$kq

Answer: $\overrightarrow{\mathrm{AF}}$ = (1 - k)p $\frac{1}{4}$kq

hence shown

How to solve vectors

How to solve vectors with ratios