# Find the angles made by the vectors displaystyle{A}={5}{i}-{2}{j}+{3}{k} with the axes give a full correct answer

permaneceerc 2020-11-20 Answered
Find the angles made by the vectors $A=5i-2j+3k$ with the axes give a full correct answer
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## Expert Answer

izboknil3
Answered 2020-11-21 Author has 99 answers
First note that the cosines of the angles a vector makes with the Cartesian coordinate axes are called the direction cosines.
Now let the given vector is
$A={A}_{x}\stackrel{^}{i}+{A}_{y}\stackrel{^}{j}+{A}_{z}\stackrel{^}{k}$
Let the vector A makes andgle $\alpha ,\beta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\gamma$ with cordinate axes x, y and z, respectively.
Also, let l, m and n are the direction cosine along x-axis, y-axis and z-axis, respectively.
Then we have
$l=\mathrm{cos}\alpha =\frac{{A}_{x}}{\frac{{\left({A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}\right)}^{1}}{2}}$
$m=\mathrm{cos}\beta =\frac{{A}_{y}}{\frac{{\left({A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}\right)}^{1}}{2}}$
$n=\mathrm{cos}\gamma =\frac{{A}_{z}}{\frac{{\left({A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}\right)}^{1}}{2}}$
It is given that
${A}_{x}=5,{A}_{y}=-2,{A}_{z}=3$
Thus we have
$\frac{{\left({A}_{x}^{2}+{A}_{y}^{2}+{A}_{z}^{2}\right)}^{1}}{2}=\sqrt{{5}^{2}+{\left(-2\right)}^{2}+{3}^{2}}=\sqrt{38}$
Thus angle with x-axis is
$\alpha ={\mathrm{cos}}^{-1}\left(\frac{5}{\sqrt{38}}\right)={35.78}^{\circ }$
The angle with y-axis is
$\beta ={\mathrm{cos}}^{-1}\left(\frac{-1}{\sqrt{38}}\right)={108.93}^{\circ }$
The angle with z-axis is
$\gamma ={\mathrm{cos}}^{-1}\left(\frac{3}{\sqrt{38}}\right)={60.88}^{\circ }$
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