Question

Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ, and the fact that a overscript right-arrow endscripts times b overscript right-arrow endscripts = axbx + ayby + azbz to calculate the angle between the two vectors given by a overscript right-arrow endscripts equals 3.0 i overscript ̂ endscripts plus 3.0 j overscript ̂ endscripts plus 3.0 k overscript ̂ endscripts and b overscript right-arrow endscripts equals 5.0 i overscript ̂ endscripts plus 7.0 j overscript ̂ endscripts plus 6.0 k overscript ̂ endscripts.

Answer 1

Answer: $\theta=cos^{-1}0.991=7.69^o$

The following vectors have been given: $\vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}$

The angle between these two vectors can be found by:

$cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\ ||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}$

$\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54$

$||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}$

$cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o$