Flashcards · 10 cards

Vectors

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Rule
The magnitude of $x\mathbf i + y\mathbf j$ ?
Answer
$\sqrt{x^2 + y^2}$ .
Magnitude|xi + yj| = √(x² + y²)
From the lesson: Vectors in components, magnitudes and bearings
Rule
A unit vector in the direction of $\mathbf a$ ?
Answer
$\hat{\mathbf a} = \dfrac{\mathbf a}{|\mathbf a|}$ : the same direction, with length 1.
A unit vectorâ = a ÷ |a| points the same way, with length 1
From the lesson: Vectors in components, magnitudes and bearings
Rule
A vector of magnitude $r$ on a bearing $\theta$ , in components?
Answer
$r\sin\theta\,\mathbf i + r\cos\theta\,\mathbf j$ , with $\mathbf i$ east and $\mathbf j$ north.
From a bearing to components(r, θ) = r sin θ i + r cos θ j
From the lesson: Vectors in components, magnitudes and bearings
Rule
$\overrightarrow{AB}$ from the position vectors $\mathbf a$ and $\mathbf b$ ?
Answer
$\mathbf b - \mathbf a$ : end minus start.
From A to BAB = b − a: end minus start
From the lesson: Position vectors and geometry
Rule
The fourth vertex $D$ of the parallelogram $ABCD$ ?
Answer
$\mathbf d = \mathbf a + \mathbf c - \mathbf b$ : go round the shape in order.
The fourth vertexd = a + c − b: go round the shape in order
From the lesson: Position vectors and geometry
Rule
The position vector of the point dividing $AB$ in the ratio $m : n$ ?
Answer
$\dfrac{n\mathbf a + m\mathbf b}{m + n}$ .
Dividing AB in the ratio 2 : 1OM = (n a + m b) ÷ (m + n)
From the lesson: Position vectors and geometry
Rule
The scalar product, and the angle between two vectors?
Answer
$\mathbf a \cdot \mathbf b = a_1b_1 + a_2b_2 = |\mathbf a||\mathbf b|\cos\theta$ , so $\cos\theta = \dfrac{\mathbf a \cdot \mathbf b}{|\mathbf a||\mathbf b|}$ .
The angle between two vectorscos θ = a·b ÷ (|a||b|)
From the lesson: The scalar product
Know it
Finding the angle $B\hat AC$ of a triangle with vectors: which two vectors?
Answer
$\overrightarrow{AB}$ and $\overrightarrow{AC}$ , both starting at $A$ .
From the lesson: The scalar product
Which method?
WAEC 2016 · Paper 2 · Q15 (a)
Given that $\mathbf m = 6\mathbf i + 8\mathbf j$ and $\mathbf n = -8\mathbf i + \frac73\mathbf j$ , find, correct to two decimal places, the magnitudes and directions (bearings) of $\mathbf m$ and $\mathbf n$ .
How do you find the bearing of $6\mathbf i + 8\mathbf j$ ?
Answer
It points 6 east and 8 north, so the angle from north is $\tan^{-1}\frac68 = 36.87^\circ$ : a bearing of $036.87^\circ$ .
From the lesson: Vectors in components, magnitudes and bearings Try the question
Which method?
WAEC 2018 · Paper 2 · Q14
The position vectors of points $A$ , $B$ and $C$ with respect to the origin are $(8\mathbf i - 10\mathbf j)$ , $(2\mathbf i + 6\mathbf j)$ and $(-10\mathbf i + 4\mathbf j)$ respectively. If $ABCN$ is a parallelogram, find:
the position vector of $N$ ;
$|\overrightarrow{AN}|$ and $|\overrightarrow{AB}|$ ;
correct to two decimal places, the acute angle between $\overrightarrow{AN}$ and $\overrightarrow{AB}$ .
How do you find $N$ in the parallelogram $ABCN$ ?
Answer
Opposite sides are equal vectors: $\overrightarrow{AB} = \overrightarrow{NC}$ , so $\mathbf n = \mathbf a + \mathbf c - \mathbf b$ .
From the lesson: Position vectors and geometry Try the question