Question

In this question, the horizontal unit vectors i and j are directed due east and due north respectively.
Find the magnitude and bearing of these vectors.
$2i+3j$
$4i-j$
$-3i+2j$
$-2i-j$

Answer 1

From the given we have been asked to find the magnitude and the bearing or simply angle made by the vectors. So, for solving these type of questions we will use the magnitude formula of vectors which is $\Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}$ and we will use the basic mathematical operation which is division in the simplification process and find the solution to the given question.

Complete step by step solution:
Firstly, the general form of a vector will be ${{a}_{x}}i+{{b}_{y}}j$.
The formulae for the magnitude of a vector in the general form will be as follows $\Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}$.
For finding the bearing or the angle the general form of the vector will be as follows.
$\Rightarrow m\sin bi+m\cos bj$
Where m is the magnitude of the vector and b is the bearing of the vector.
So, for the case $1$ that is for the $2i+3j$, the magnitude will be as follows.
$\Rightarrow \sqrt{{{2}^{2}}+{{3}^{2}}}$
$\Rightarrow \sqrt{13}$
So, we get that when we divide the coefficient of I and j in $\Rightarrow m\cos bi+m\sin bj$ we get,
$\Rightarrow m\cos bi+m\sin bj$
So, we get for our case as follows.
$\Rightarrow \cot b=\dfrac{2}{3}$
$\Rightarrow \cot b=\dfrac{2}{3}={{33.7}^{\circ }}$
The $2i+3j$ is in first quadrant so, bearing will be $\Rightarrow {{33.7}^{\circ }}\approx {{34}^{^{\circ }}}$
For the second case that is for $4i-j$ we get the results as follows.
Magnitude will be as follows.
$\Rightarrow \sqrt{{{4}^{2}}+{{(-1)}^{2}}}$
$\Rightarrow \sqrt{17}$
The bearing will be as follows.
$\Rightarrow \cot b=-4={{14}^{\circ }}$
The vector is in the fourth quadrant, so we get a bearing as follows.
$\Rightarrow {{90}^{\circ }}+{{14}^{\circ }}={{100}^{\circ }}$
For the third case that is for $-3i+2j$ we get the results as follows.
Magnitude will be as follows.
$\Rightarrow \sqrt{{{2}^{2}}+{{(-3)}^{2}}}$
$\Rightarrow \sqrt{13}$
The bearing will be as follows.
$\Rightarrow \cot b=\dfrac{-3}{2}={{33.7}^{\circ }}$
This is in second quadrant so bearing will be $\Rightarrow {{270}^{\circ }}+{{33.7}^{\circ }}\approx {{304}^{\circ }}$.
For the fourth case that is for $-2i-j$ we get the results as follows.
Magnitude will be as follows.
$\Rightarrow \sqrt{{{(-2)}^{2}}+{{(-1)}^{2}}}$
$\Rightarrow \sqrt{5}$
The bearing will be as follows.
$\Rightarrow \cot b=2={{26.6}^{\circ }}$
This is in third quadrant so bearing will be $\Rightarrow {{270}^{\circ }}-{{26.6}^{\circ }}\approx {{243}^{\circ }}$.

Note: Students should do the calculations very carefully. Students should have good knowledge in the concept of vectors and also in the concept quadrants in trigonometry. Students should know the formulae in vectors like.
Magnitude formula of vectors which is $\Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}$
One of the general forms of vectors $\Rightarrow m\sin bi+m\cos bj$.