Q123 / 160 · AP EAMCET 2022 (05 Jul Shift 1) · Physics
The component of a vector $\mathbf{P}=3 \hat{i}+4 \hat{j}$ alog the direction $(\hat{i}+2 \hat{j})$ is
- A$\frac{8}{\sqrt{5}}$
- Correct answerB$\frac{11}{\sqrt{5}}$
- C$\frac{11}{2}$
- D$\sqrt{10}$
Solution
Given, $\mathbf{P}=3 \hat{i}+4 \hat{j}$
Let $Q=\hat{i}+2 \hat{j}$

The component of $\mathbf{P}$ along the $\mathbf{Q}$ is given as $|\mathbf{P}| \cos \theta$ Hence, we know that, $\mathbf{P} \cdot \mathbf{Q}=|\mathbf{P}||\mathbf{Q}| \cos \theta$ $\Rightarrow|P| \cos \theta=\frac{\mathbf{P} \cdot \mathbf{Q}}{|\mathbf{Q}|}$ $=\frac{(3 \hat{i}+4 \hat{j}) \cdot(\hat{i}+2 \hat{j})}{\sqrt{1^2+2^2}}=\frac{3+8}{\sqrt{5}}=\frac{11}{\sqrt{5}}$

The component of $\mathbf{P}$ along the $\mathbf{Q}$ is given as $|\mathbf{P}| \cos \theta$ Hence, we know that, $\mathbf{P} \cdot \mathbf{Q}=|\mathbf{P}||\mathbf{Q}| \cos \theta$ $\Rightarrow|P| \cos \theta=\frac{\mathbf{P} \cdot \mathbf{Q}}{|\mathbf{Q}|}$ $=\frac{(3 \hat{i}+4 \hat{j}) \cdot(\hat{i}+2 \hat{j})}{\sqrt{1^2+2^2}}=\frac{3+8}{\sqrt{5}}=\frac{11}{\sqrt{5}}$
Hence, the correct answer is option: 'B'.