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Naya [18.7K]
3 years ago
10

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.
Physics
1 answer:
makkiz [27]3 years ago
7 0

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

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