If the magnitude of two vectors are 4 unit and 6 unit and magnitude of their scalar product is 12√2 unit. What is the angle between the vectors?

Solution:

Let the two vectors be $\vec{A}$ and $\vec{B}$.

Given:

Magnitude of vector A, $|\vec{A}| = 4$ unit

Magnitude of vector B, $|\vec{B}| = 6$ unit

Magnitude of their scalar product (dot product), $\vec{A} \cdot \vec{B} = 12\sqrt{2}$ unit

The scalar product of two vectors is given by the formula:

$\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos\theta$

where $\theta$ is the angle between the two vectors.

Substitute the given values into the formula:

$12\sqrt{2} = (4)(6) \cos\theta$

$12\sqrt{2} = 24 \cos\theta$

Now, solve for $\cos\theta$:

$\cos\theta = \frac{12\sqrt{2}}{24}$

$\cos\theta = \frac{\sqrt{2}}{2}$

$\cos\theta = \frac{1}{\sqrt{2}}$

To find the angle $\theta$, we take the inverse cosine:

$\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$\theta = 45^\circ$

Thus, the angle between the vectors is 45°.