If the angle between the two vectors vec{u} = | vedclass

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(D) Given vectors are $\vec{u} = \hat{i} + \hat{k}$ and $\vec{v} = \hat{i} - \hat{j} + a\hat{k}$.The angle $	heta$ between two vectors is given by $\

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$0$

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(D) Given vectors are $\vec{u} = \hat{i} + \hat{k}$ and $\vec{v} = \hat{i} - \hat{j} + a\hat{k}$.The angle $	heta$ between two vectors is given by $\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$.Here,$	heta = \pi/3$,so $\cos(\pi/3) = 1/2$.The dot product $\vec{u} \cdot \vec{v} = (1)(1) + (0)(-1) + (1)(a) = 1 + a$.The magnitudes are $|\vec{u}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$ and $|\vec{v}| = \sqrt{1^2 + (-1)^2 + a^2} = \sqrt{2 + a^2}$.Substituting these into the formula: $\frac{1}{2} = \frac{1 + a}{\sqrt{2} \sqrt{2 + a^2}}$.Squaring both sides: $\frac{1}{4} = \frac{(1 + a)^2}{2(2 + a^2)} \implies \frac{1}{2} = \frac{1 + 2a + a^2}{2 + a^2}$.Cross-multiplying: $2 + a^2 = 2(1 + 2a + a^2) = 2 + 4a + 2a^2$.Rearranging terms: $a^2 + 4a = 0 \implies a(a + 4) = 0$.Thus,$a = 0$ or $a = -4$. Given the options,the correct value is $a = 0$.

If the angle between the two vectors $\vec{u} = \hat{i} + \hat{k}$ and $\vec{v} = \hat{i} - \hat{j} + a\hat{k}$ is $\pi/3$,find the value of $a$.

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