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(A) Let $\vec{a}$ make an angle $\alpha$ with the $Z$-axis and $\vec{b}$ make an angle $\beta$ with the $X$-axis. Using the direction cosines property

Vedclass · Accepted Answer

$(1-\sqrt{2}) \hat{i}-\hat{j}+\hat{k}$

Vedclass · Answer

(A) Let $\vec{a}$ make an angle $\alpha$ with the $Z$-axis and $\vec{b}$ make an angle $\beta$ with the $X$-axis. Using the direction cosines property $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$:For vector $\vec{a}$: $\cos^2 60^{\circ} + \cos^2 60^{\circ} + \cos^2 \alpha = 1 \Rightarrow \frac{1}{4} + \frac{1}{4} + \cos^2 \alpha = 1 \Rightarrow \cos^2 \alpha = \frac{1}{2} \Rightarrow \cos \alpha = \pm \frac{1}{\sqrt{2}}$.Thus, $\vec{a} = 2(\cos 60^{\circ} \hat{i} + \cos 60^{\circ} \hat{j} + \cos \alpha \hat{k}) = 2(\frac{1}{2} \hat{i} + \frac{1}{2} \hat{j} \pm \frac{1}{\sqrt{2}} \hat{k}) = \hat{i} + \hat{j} \pm \sqrt{2} \hat{k}$.For vector $\vec{b}$: $\cos^2 \beta + \cos^2 45^{\circ} + \cos^2 45^{\circ} = 1 \Rightarrow \cos^2 \beta + \frac{1}{2} + \frac{1}{2} = 1 \Rightarrow \cos \beta = 0$.Thus, $\vec{b} = \sqrt{2}(0 \hat{i} + \cos 45^{\circ} \hat{j} + \cos 45^{\circ} \hat{k}) = \sqrt{2}(0 \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}) = \hat{j} + \hat{k}$.Now, $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & \pm \sqrt{2} \ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - (\pm \sqrt{2})) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = (1 \mp \sqrt{2}) \hat{i} - \hat{j} + \hat{k}$.Taking the positive sign, we get $(1 - \sqrt{2}) \hat{i} - \hat{j} + \hat{k}$.

$A$ vector $\vec{a}$ of length $2$ units makes an angle $60^{\circ}$ with each of the $X$-axis and $Y$-axis. If another vector $\vec{b}$ of length $\sqrt{2}$ units makes an angle $45^{\circ}$ with each of the $Y$-axis and $Z$-axis, then $\vec{a} \times \vec{b} = $

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