Let bar{a} = a_1 hat{i} + a_2 hat{j} + a_3 ha | vedclass

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(A) Given $\bar{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\bar{b} = \sqrt{2} \hat{i} + 3 \sqrt{2} \hat{j} + 4 \hat{k}$.We have $|\bar{b}| = \

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$XY$-plane

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(A) Given $\bar{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\bar{b} = \sqrt{2} \hat{i} + 3 \sqrt{2} \hat{j} + 4 \hat{k}$.We have $|\bar{b}| = \sqrt{(\sqrt{2})^2 + (3 \sqrt{2})^2 + 4^2} = \sqrt{2 + 18 + 16} = \sqrt{36} = 6$.The angle between $\bar{a}$ and $\bar{b}$ is $45^{\circ}$,so $\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos(45^{\circ})$.$\bar{a} \cdot \bar{b} = a_1(\sqrt{2}) + a_2(3 \sqrt{2}) + a_3(4) = \sqrt{2}(a_1 + 3a_2) + 4a_3$.Thus,$\sqrt{2}(a_1 + 3a_2) + 4a_3 = |\bar{a}| \cdot 6 \cdot \frac{1}{\sqrt{2}} = 3 \sqrt{2} |\bar{a}|$.Rearranging,we get $\sqrt{2}(a_1 + 3a_2 - 3|\bar{a}|) + 4a_3 = 0$.Since $a_1, a_2, a_3$ and $|\bar{a}|$ are rational,for the equation to hold,the irrational part must be zero and the rational part must be zero.Thus,$4a_3 = 0 \Rightarrow a_3 = 0$.Since $a_3 = 0$,the vector $\bar{a} = a_1 \hat{i} + a_2 \hat{j}$ lies in the $XY$-plane.

Let $\bar{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$,where $a_1, a_2, a_3$ and $|\bar{a}|$ are rational numbers. If $\bar{a}$ makes an angle of $45^{\circ}$ with $\bar{b} = \sqrt{2} \hat{i} + 3 \sqrt{2} \hat{j} + 4 \hat{k}$,then $\bar{a}$ lies in:

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