Q61 / 155 · WBJEE 2024 · Mathematics
A unit vector in $X Y$-plane making an angle $45^{\circ}$ with $\hat{i}+\hat{j}$ and an angle $60^{\circ}$ with $3 \hat{i}-4 \hat{j}$ is
- Correct answerA$\frac{13}{14} \hat{i}+\frac{1}{14} \hat{j}$
- B$\frac{1}{14} \hat{i}+\frac{13}{14} \hat{j}$
- C$\frac{13}{14} \hat{i}-\frac{1}{14} \hat{j}$
- D$\frac{1}{14} \hat{i}-\frac{13}{14} \hat{j}$
Solution
Hint: Let the vector be $x \hat{i}+y \hat{j}=\vec{r} \quad \vec{a}=\hat{i}+\hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}$
$\begin{aligned}
& \vec{r} \cdot \vec{a}=|\vec{r}||\vec{a}| \cos 45^{\circ}=1 \cdot \sqrt{2} \cdot \frac{1}{\sqrt{2}}=1 \quad \Rightarrow x+y=1 \cdots-\cdots(1) \\
& \vec{r} \cdot \vec{b}=|\vec{r}||\vec{b}| \cos 60^{\circ} \Rightarrow 3 x-4 y=1 \cdot 5 \cdot \frac{1}{2} \Rightarrow \frac{2}{5}(3 x-4 y)=1 \Rightarrow 6 x-8 y=5 ...(2)
\end{aligned}$
Solving (1) and (2), $14 \mathrm{x}=13 \Rightarrow \mathrm{x}=\frac{13}{14} \quad \therefore \mathrm{y}=\frac{1}{14}$
$\therefore \quad$ Unit vector $=\frac{13}{14} \hat{i}+\frac{1}{14} \hat{j}$
Hence, the correct answer is option: 'A'.