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Worked solution

Prove the directional-cosine identity for a position vector

OA→=i+4j+7k\overrightarrow{OA}=\mathbf i+4\mathbf j+7\mathbf k makes angles θ,ϕ,ω\theta,\phi,\omega with the positive x-, y-, z-axes respectively. Show that cos⁡2θ+cos⁡2ϕ+cos⁡2ω=1\cos^2\theta+\cos^2\phi+\cos^2\omega=1.

CJC · Vectors Basic · Example 14

Answer

cos⁡θ=1/66\cos\theta=1/\sqrt{66}, cos⁡ϕ=4/66\cos\phi=4/\sqrt{66}, cos⁡ω=7/66\cos\omega=7/\sqrt{66}; the sum of their squares is 11.

Compare your reasoning

Use the positive x-axis unit vector

cos⁡θ=OA→⋅(100)∣OA→∣∣(100)∣=(147)⋅(100)∣(147)∣∣(100)∣=166\cos\theta=\frac{\overrightarrow{OA}\cdot\begin{pmatrix}1\\0\\0\end{pmatrix}}{|\overrightarrow{OA}|\left|\begin{pmatrix}1\\0\\0\end{pmatrix}\right|}=\frac{\begin{pmatrix}1\\4\\7\end{pmatrix}\cdot\begin{pmatrix}1\\0\\0\end{pmatrix}}{\left|\begin{pmatrix}1\\4\\7\end{pmatrix}\right|\left|\begin{pmatrix}1\\0\\0\end{pmatrix}\right|}=\frac{1}{\sqrt{66}}

The unit direction has magnitude one; the common position-vector magnitude is √66.

Use the positive y-axis unit vector

cos⁡ϕ=OA→⋅(010)∣OA→∣∣(010)∣=(147)⋅(010)∣(147)∣∣(010)∣=466\cos\phi=\frac{\overrightarrow{OA}\cdot\begin{pmatrix}0\\1\\0\end{pmatrix}}{|\overrightarrow{OA}|\left|\begin{pmatrix}0\\1\\0\end{pmatrix}\right|}=\frac{\begin{pmatrix}1\\4\\7\end{pmatrix}\cdot\begin{pmatrix}0\\1\\0\end{pmatrix}}{\left|\begin{pmatrix}1\\4\\7\end{pmatrix}\right|\left|\begin{pmatrix}0\\1\\0\end{pmatrix}\right|}=\frac{4}{\sqrt{66}}

The unit direction has magnitude one; the common position-vector magnitude is √66.

Use the positive z-axis unit vector

cos⁡ω=OA→⋅(001)∣OA→∣∣(001)∣=(147)⋅(001)∣(147)∣∣(001)∣=766\cos\omega=\frac{\overrightarrow{OA}\cdot\begin{pmatrix}0\\0\\1\end{pmatrix}}{|\overrightarrow{OA}|\left|\begin{pmatrix}0\\0\\1\end{pmatrix}\right|}=\frac{\begin{pmatrix}1\\4\\7\end{pmatrix}\cdot\begin{pmatrix}0\\0\\1\end{pmatrix}}{\left|\begin{pmatrix}1\\4\\7\end{pmatrix}\right|\left|\begin{pmatrix}0\\0\\1\end{pmatrix}\right|}=\frac{7}{\sqrt{66}}

The unit direction has magnitude one; the common position-vector magnitude is √66.

Square each cosine before adding

cos⁡2θ+cos⁡2ϕ+cos⁡2ω=166+1666+4966=1\begin{aligned}&\cos^2\theta+\cos^2\phi+\cos^2\omega\\ & =\frac1{66}+\frac{16}{66}+\frac{49}{66}\\ & =1\end{aligned}

The three squared components of a unit direction add to one.

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Solution notes

Both copies print the full three-axis substitution proof.