Question

The magnitude of component of vector A = $3\hat{i} + 4\hat{j}$ along vector $2\hat{i} + \hat{j}$ is $\alpha$. Then find the value of $\frac{\alpha^2}{4}$.

Answer 1

5

Answer 2

Let $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = 2\hat{i} + \hat{j}$.

The magnitude of the component of vector $\vec{A}$ along vector $\vec{B}$ is given by the scalar projection of $\vec{A}$ onto $\vec{B}$, which is $\alpha = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.

First, calculate the dot product $\vec{A} \cdot \vec{B}$: $\vec{A} \cdot \vec{B} = (3)(2) + (4)(1) = 6 + 4 = 10$.

Next, calculate the magnitude of vector $\vec{B}$: $|\vec{B}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$.

Now, calculate the value of $\alpha$: $\alpha = \frac{10}{\sqrt{5}}$.

We are asked to find the value of $\frac{\alpha^2}{4}$. Calculate $\alpha^2$: $\alpha^2 = \left(\frac{10}{\sqrt{5}}\right)^2 = \frac{10^2}{(\sqrt{5})^2} = \frac{100}{5} = 20$.

Finally, calculate $\frac{\alpha^2}{4}$: $\frac{\alpha^2}{4} = \frac{20}{4} = 5$.

The value of $\frac{\alpha^2}{4}$ is 5.