Question

A vector has magnitude same as that of A = $-3\hat{i} + 4\hat{j}$ and is parallel to B = $4\hat{i} + 3\hat{j}$. The x and y components of this vector in the first quadrant are x and y respectively where:

x = \\_\\_\\_\\_.

Answer 1

To find the x and y components of the vector, we first need to determine the magnitude of vector A.

Magnitude of Vector A: The magnitude |A| is calculated as:

$|A| = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$

Direction of Vector B: The unit vector in the direction of vector B is given by:

$\hat{b} = \frac{\mathbf{B}}{|\mathbf{B}|}.$

First, we need the magnitude of vector B :

$|B| = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.$

The unit vector $\hat{b}$ is:

$\hat{b} = \frac{4\hat{i} + 3\hat{j}}{5} = \frac{4}{5}\hat{i} + \frac{3}{5}\hat{j}.$

Magnitude of New Vector: Since the new vector has the same magnitude as A and is parallel to B , we can write it as:

$\mathbf{V} = k \hat{b} = k \left(\frac{4}{5}\hat{i} + \frac{3}{5}\hat{j}\right).$

To find k , we set the magnitude of V to be equal to the magnitude of A :

$|V| = k \times |\hat{b}| = k \times 1 = k.$

Thus, we have:

$k = 5.$

Finding Components: The components of vector V :

$x = 5 \times \frac{4}{5} = 4,$

$y = 5 \times \frac{3}{5} = 3.$