Question

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

Answer 1

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
\(\displaystyle{A}={A}_{{x}}\hat{{i}}+{A}_{{y}}\hat{{j}}+{A}_{{z}}\hat{{k}}\)
Let the vector A makes andgle \(\displaystyle\alpha,\beta{\quad\text{and}\quad}\gamma\) 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
\(\displaystyle{l}= \cos{\alpha}=\frac{{{A}_{{x}}}}{{\frac{{\left({{A}_{{x}}^{{2}}}+{{A}_{{y}}^{{2}}}+{{A}_{{z}}^{{2}}}\right)}^{1}}{{2}}}}\)
\(\displaystyle{m}= \cos{\beta}=\frac{{{A}_{{y}}}}{{\frac{{\left({{A}_{{x}}^{{2}}}+{{A}_{{y}}^{{2}}}+{{A}_{{z}}^{{2}}}\right)}^{1}}{{2}}}}\)
\(\displaystyle{n}= \cos{\gamma}=\frac{{{A}_{{z}}}}{{\frac{{\left({{A}_{{x}}^{{2}}}+{{A}_{{y}}^{{2}}}+{{A}_{{z}}^{{2}}}\right)}^{1}}{{2}}}}\)
It is given that
\(\displaystyle{A}_{{x}}={5},{A}_{{y}}=-{2},{A}_{{z}}={3}\)
Thus we have
\(\displaystyle\frac{{\left({{A}_{{x}}^{{2}}}+{{A}_{{y}}^{{2}}}+{{A}_{{z}}^{{2}}}\right)}^{1}}{{2}}=\sqrt{{{5}^{2}+{\left(-{2}\right)}^{2}+{3}^{2}}}=\sqrt{{38}}\)
Thus angle with x-axis is
\(\displaystyle\alpha={{\cos}^{{-{1}}}{\left(\frac{5}{\sqrt{{38}}}\right)}}={35.78}^{\circ}\)
The angle with y-axis is
\(\displaystyle\beta={{\cos}^{ -{{1}}}{\left(\frac{{-{1}}}{\sqrt{{38}}}\right)}}={108.93}^{\circ}\)
The angle with z-axis is
\(\displaystyle\gamma={{\cos}^{ -{{1}}}{\left(\frac{3}{\sqrt{{38}}}\right)}}={60.88}^{\circ}\)