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

\(\displaystyle\gamma={{\cos}^{ -{{1}}}{\left(\frac{3}{\sqrt{{38}}}\right)}}={60.88}^{\circ}\)

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

\(\displaystyle\gamma={{\cos}^{ -{{1}}}{\left(\frac{3}{\sqrt{{38}}}\right)}}={60.88}^{\circ}\)