Question

If $\bar{a}$ = î - 2ĵ + 3k and $\bar{b}$ = 2î + 3ĵ − k are two vectors, then the angle between the vectors 3$\bar{a}$ + 5$\bar{b}$ and 5$\bar{a}$ + 3$\bar{b}$ is

• A. $\cos^{-1}(\frac{10}{19})$
• B. $\cos^{-1}(\frac{11}{19})$
• C. $\cos^{-1}(\frac{13}{19})$
• D. $\cos^{-1}(\frac{14}{19})$

Answer 1

Answer: (C)

$\cos^{-1}(\frac{13}{19})$

Answer 2

Let

$\mathbf{a} = \langle 1, -2, 3 \rangle, \quad \mathbf{b} = \langle 2, 3, -1 \rangle$.

Compute

$\mathbf{u} = 3\mathbf{a} + 5\mathbf{b} = 3\langle 1,-2,3 \rangle + 5\langle 2,3,-1 \rangle = \langle 3+10,\,-6+15,\,9-5 \rangle = \langle 13, 9, 4 \rangle$,

$\mathbf{v} = 5\mathbf{a} + 3\mathbf{b} = 5\langle 1,-2,3 \rangle + 3\langle 2,3,-1 \rangle = \langle 5+6,\,-10+9,\,15-3 \rangle = \langle 11, -1, 12 \rangle$.

Now, find the dot product:

$\mathbf{u}\cdot\mathbf{v} = (13)(11) + (9)(-1) + (4)(12) = 143 - 9 + 48 = 182$.

Determine the magnitudes:

$|\mathbf{u}|=\sqrt{13^2+9^2+4^2}=\sqrt{169+81+16}=\sqrt{266}$,

$|\mathbf{v}|=\sqrt{11^2+(-1)^2+12^2}=\sqrt{121+1+144}=\sqrt{266}$.

Thus, the cosine of the angle $\theta$ between $\mathbf{u}$ and $\mathbf{v}$ is

$\cos \theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|} = \frac{182}{266} = \frac{91}{133} = \frac{13}{19}$.

Hence,

$\theta = \cos^{-1}\left(\frac{13}{19}\right)$.