Porisma

Worked solution

Keep the sign when finding the angle between vectors

Find the angle between a=(122)\mathbf a=\begin{pmatrix}1\\2\\2\end{pmatrix} and b=(13−4)\mathbf b=\begin{pmatrix}1\\3\\-4\end{pmatrix}.

CJC · Vectors Basic · Example 12

Answer

θ=cos⁡−1 ⁣(−1326)=93.7∘\theta=\cos^{-1}\!\left(-\dfrac1{3\sqrt{26}}\right)=93.7^\circ (1 d.p.), or 1.641.64 rad (3 s.f.; corrected).

Compare your reasoning

Evaluate the dot product and lengths

a⋅b=∣a∣∣b∣cos⁡θ,(122)⋅(13−4)=∣(122)∣∣(13−4)∣cos⁡θ,1+6−8=926cos⁡θ,cos⁡θ=−1326.\begin{aligned}\mathbf a\cdot\mathbf b&=|\mathbf a||\mathbf b|\cos\theta,\\\begin{pmatrix}1\\2\\2\end{pmatrix}\cdot\begin{pmatrix}1\\3\\-4\end{pmatrix}&=\left|\begin{pmatrix}1\\2\\2\end{pmatrix}\right|\left|\begin{pmatrix}1\\3\\-4\end{pmatrix}\right|\cos\theta,\\1+6-8&=\sqrt9\sqrt{26}\cos\theta,\\\cos\theta&=-\frac1{3\sqrt{26}}.\end{aligned}

Let θ be the angle between the two nonzero vectors. Keep the negative dot product.

Take inverse cosine before rounding

θ=cos⁡−1 ⁣(−1326)≈93.7∘ (1 d.p.)≈1.64 rad (3 s.f.)\begin{aligned}&\theta\\ & =\cos^{-1}\!\left(-\frac1{3\sqrt{26}}\right)\approx93.7^\circ\text{ (1 d.p.)}\approx1.64\text{ rad (3 s.f.)}\end{aligned}

The exact value is about 1.636215 rad or 93.748215 degrees. A negative cosine makes this an obtuse angle.

How did this question go?

Common mistakes

Taking the absolute value of the dot product here would give the supplementary acute angle, not the requested angle between these directed vectors.

Solution notes

The Teacher prints 93.7° (to 1 d.p.) or 1.63 rad (to 3 s.f.). The degree value is correct; the radian rounding should be 1.64, since the exact angle is about 1.636215 rad. The printed 1.63 is preserved in the unchanged native transcription and review evidence.