A unit vector in XY-plane making an angle 45^ | vedclass

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(A) Let the unit vector be $\vec{r} = x\hat{i} + y\hat{j}$. Since it is a unit vector, $x^2 + y^2 = 1$. Given $\vec{a} = \hat{i} + \hat{j}$ and $\vec{

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$\frac{13}{14}\hat{i}+\frac{1}{14}\hat{j}$

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(A) Let the unit vector be $\vec{r} = x\hat{i} + y\hat{j}$. Since it is a unit vector, $x^2 + y^2 = 1$. Given $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = 3\hat{i} - 4\hat{j}$. The angle between $\vec{r}$ and $\vec{a}$ is $45^{\circ}$, so $\vec{r} \cdot \vec{a} = |\vec{r}||\vec{a}| \cos 45^{\circ} = 1 \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$. Thus, $x + y = 1 \implies y = 1 - x$. The angle between $\vec{r}$ and $\vec{b}$ is $60^{\circ}$, so $\vec{r} \cdot \vec{b} = |\vec{r}||\vec{b}| \cos 60^{\circ} = 1 \cdot \sqrt{3^2 + (-4)^2} \cdot \frac{1}{2} = 5 \cdot \frac{1}{2} = \frac{5}{2}$. Thus, $3x - 4y = \frac{5}{2} \implies 6x - 8y = 5$. Substituting $y = 1 - x$ into the second equation: $6x - 8(1 - x) = 5 \implies 6x - 8 + 8x = 5 \implies 14x = 13 \implies x = \frac{13}{14}$. Then $y = 1 - \frac{13}{14} = \frac{1}{14}$. Therefore, the unit vector is $\frac{13}{14}\hat{i} + \frac{1}{14}\hat{j}$.

$A$ unit vector in $XY$-plane making an angle $45^{\circ}$ with $\hat{i}+\hat{j}$ and an angle $60^{\circ}$ with $3\hat{i}-4\hat{j}$ is

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