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(D) Let the unit vector be $\vec{u} = x\,i + y\,j$. Since it is a unit vector,$x^2 + y^2 = 1$.Given the angle with $\vec{a} = i + j$ is $45^{\circ}$,w

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None of these

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(D) Let the unit vector be $\vec{u} = x\,i + y\,j$. Since it is a unit vector,$x^2 + y^2 = 1$.Given the angle with $\vec{a} = i + j$ is $45^{\circ}$,we have $\vec{u} \cdot \vec{a} = |\vec{u}| |\vec{a}| \cos(45^{\circ})$.$(x\,i + y\,j) \cdot (i + j) = (1)(\sqrt{1^2 + 1^2}) \frac{1}{\sqrt{2}} \Rightarrow x + y = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$.Given the angle with $\vec{b} = 3i - 4j$ is $60^{\circ}$,we have $\vec{u} \cdot \vec{b} = |\vec{u}| |\vec{b}| \cos(60^{\circ})$.$(x\,i + y\,j) \cdot (3i - 4j) = (1)(\sqrt{3^2 + (-4)^2}) \frac{1}{2} \Rightarrow 3x - 4y = 5 \cdot \frac{1}{2} = 2.5$.We have the system: $x + y = 1$ and $3x - 4y = 2.5$.From the first equation,$y = 1 - x$. Substituting into the second: $3x - 4(1 - x) = 2.5 \Rightarrow 3x - 4 + 4x = 2.5 \Rightarrow 7x = 6.5 \Rightarrow x = \frac{6.5}{7} = \frac{13}{14}$.Then $y = 1 - \frac{13}{14} = \frac{1}{14}$.Check the unit vector condition: $x^2 + y^2 = (\frac{13}{14})^2 + (\frac{1}{14})^2 = \frac{169 + 1}{196} = \frac{170}{196} 
eq 1$.Since no option satisfies these conditions,the correct answer is $(d)$.

$A$ unit vector in the $xy$-plane that makes an angle of $45^{\circ}$ with the vector $(i + j)$ and an angle of $60^{\circ}$ with the vector $(3i - 4j)$ is:

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