If vec{a} = -4 hat{i} + 2 hat{j} + 4 hat{k} a | vedclass

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(B) The angle $	heta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by $\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$.Here,

Vedclass · Accepted Answer

$135$

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(B) The angle $	heta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by $\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$.Here,$\vec{u} = 2 \vec{a}$ and $\vec{v} = \frac{\vec{b}}{2}$.Thus,$\cos 	heta = \frac{(2 \vec{a}) \cdot (\frac{\vec{b}}{2})}{|2 \vec{a}| |\frac{\vec{b}}{2}|} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.First,calculate the dot product $\vec{a} \cdot \vec{b} = (-4)(\sqrt{2}) + (2)(-\sqrt{2}) + (4)(0) = -4 \sqrt{2} - 2 \sqrt{2} = -6 \sqrt{2}$.Next,calculate the magnitudes:$|\vec{a}| = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.$|\vec{b}| = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2 + 0^2} = \sqrt{2 + 2} = \sqrt{4} = 2$.Substituting these values into the formula:$\cos 	heta = \frac{-6 \sqrt{2}}{6 	imes 2} = \frac{-6 \sqrt{2}}{12} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.Since $\cos 	heta = -\frac{1}{\sqrt{2}}$,we have $	heta = 135^{\circ}$.

If $\vec{a} = -4 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and $\vec{b} = \sqrt{2} \hat{i} - \sqrt{2} \hat{j}$ are two vectors,then the angle between the vectors $2 \vec{a}$ and $\frac{\vec{b}}{2}$ is (in $^{\circ}$)

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