If overline{a}=hat{i}-2 hat{j}+3 hat{k} and o | vedclass

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(C) Let $\overline{u} = 3 \overline{a} + 5 \overline{b}$ and $\overline{v} = 5 \overline{a} + 3 \overline{b}$.$\overline{u} = 3(\hat{i}-2 \hat{j}+3 \h

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

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(C) Let $\overline{u} = 3 \overline{a} + 5 \overline{b}$ and $\overline{v} = 5 \overline{a} + 3 \overline{b}$.$\overline{u} = 3(\hat{i}-2 \hat{j}+3 \hat{k}) + 5(2 \hat{i}+3 \hat{j}-\hat{k}) = (3+10)\hat{i} + (-6+15)\hat{j} + (9-5)\hat{k} = 13 \hat{i} + 9 \hat{j} + 4 \hat{k}$.$\overline{v} = 5(\hat{i}-2 \hat{j}+3 \hat{k}) + 3(2 \hat{i}+3 \hat{j}-\hat{k}) = (5+6)\hat{i} + (-10+9)\hat{j} + (15-3)\hat{k} = 11 \hat{i} - \hat{j} + 12 \hat{k}$.The angle $	heta$ between $\overline{u}$ and $\overline{v}$ is given by $\cos 	heta = \frac{\overline{u} \cdot \overline{v}}{|\overline{u}| |\overline{v}|}$.$\overline{u} \cdot \overline{v} = (13)(11) + (9)(-1) + (4)(12) = 143 - 9 + 48 = 182$.$|\overline{u}| = \sqrt{13^2 + 9^2 + 4^2} = \sqrt{169 + 81 + 16} = \sqrt{266}$.$|\overline{v}| = \sqrt{11^2 + (-1)^2 + 12^2} = \sqrt{121 + 1 + 144} = \sqrt{266}$.$\cos 	heta = \frac{182}{\sqrt{266} \cdot \sqrt{266}} = \frac{182}{266} = \frac{13}{19}$.Therefore,$	heta = \cos^{-1}\left(\frac{13}{19}ight)$.

If $\overline{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\overline{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ are two vectors,then the angle between the vectors $3 \bar{a}+5 \bar{b}$ and $5 \bar{a}+3 \bar{b}$ is

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