WBJEE 2024 Math Paper

Let's denote the unit vector as $\vec{u}=a\hat{i}+b\hat{j}$, where $a$ and $b$ are the components of the vector along the $x$ and $y$ axes, respectively. According to the given problem, this unit vector makes an angle of $45^{\circ}$ with $\hat{i}+\hat{j}$ and an angle of $60^{\circ}$ with $3\hat{i}-4\hat{j}$.We can start by using the dot product formula to generate necessary equations. The dot product of two vectors $\vec{u}$ and $\vec{v}$ is given by: $\vec{u}\cdot\vec{v}=|\vec{u}||\vec{v}|\cos\theta$Since $\vec{u}$ is a unit vector, $|\vec{u}|=1$, and thus the dot product simplifies to:$\vec{u}\cdot\vec{v}=|\vec{v}|\cos\theta$Let's apply this to both given vectors. For the first vector, $\hat{i}+\hat{j}$ :Here, $\vec{v}_1=\hat{i}+\hat{j}$ and the angle is $45^{\circ}$.The magnitude of $\vec{v}_1$ is:$|\vec{v}_1|=\sqrt{1^2+1^2}=\sqrt{2}$The equation becomes:$\;\vec{u}\cdot(\hat{i}+\hat{j})=\sqrt{2}\cos45^{\circ}$$\;a+b=\sqrt{2}\cdot\;\frac{1}{\sqrt{2}}$$\;a+b=1$ For the second vector, $3\hat{i}-4\hat{j}$ :Here, $\vec{v}_2=3\hat{i}-4\hat{j}$ and the angle is $60^{\circ}$.The magnitude of $\vec{v}_2$ is:$|\vec{v}_2|=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5$The equation becomes:$\;\vec{u}\cdot(3\hat{i}-4\hat{j})=5\cos60^{\circ}$$\;3a-4b=5\cdot\;\frac{1}{2}$$\;3a-4b=\;\frac{5}{2}$ Now we have a system of two equations:$\;a+b=1$$\;3a-4b=\;\frac{5}{2}$We can solve this system of equations to find the values of $a$ and $b$.From the first equation, solve for $b$ :$b=1-a$Substitute this into the second equation: $\;3a-4(1-a)=\;\frac{5}{2}$$\;3a-4+4a=\;\frac{5}{2}$$\;7a-4=\;\frac{5}{2}$$\;7a=\;\frac{5}{2}+4$$\;7a=\;\frac{5}{2}+\;\frac{8}{2}$$\;7a=\;\frac{13}{2}$$\;a=\;\frac{13}{14}$ Substitute back to find $b$ :$\;b=1-\;\frac{13}{14}$$\;b=\;\frac{14}{14}-\;\frac{13}{14}$$\;b=\;\frac{1}{14}$Thus, the unit vector is:$\vec{u}=\;\frac{13}{14}\hat{i}+\;\frac{1}{14}\hat{j}$Therefore, the correct option is:Option A: $\;\frac{13}{14}\hat{i}+\;\frac{1}{14}\hat{j}$But $|\vec{u}|=\sqrt{ \left(\;\frac{13}{14}\right)^2+ \left(\;\frac{1}{14}\right)^2} \neq 1$. So, $\vec{u}$ is not an unit vector.