Question

\( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \(|\vec{a}| = 3\), \(|\vec{b}| = 2\sqrt{2}\), \(|\vec{c}| = 5\), and \( \vec{c} \) is perpendicular to the plane of \( \vec{a} \) and \( \vec{b} \).

If the angle between the vectors \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{4} \), then

\[ |\vec{a} + \vec{b} + \vec{c}| = \ ? \]

• A. \(5\sqrt{3}\)
• B. \(2\sqrt{5}\)
• C. \(10\)
• D. \(3\sqrt{6}\)

Hint:

When vectors are perpendicular or in specific geometric arrangements, utilize geometric relationships and trigonometric identities to simplify magnitude calculations.

Answer 1

The Correct Option is D

Using the given magnitudes and the angle between $\vec{a}$ and $\vec{b}$, the magnitude of the sum $\vec{a} + \vec{b}$ in the plane can be calculated using the cosine rule for vectors: \[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos\left(\frac{\pi}{4}\right) \] \[ |\vec{a} + \vec{b}|^2 = 3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2} \times \frac{\sqrt{2}}{2} \] \[ |\vec{a} + \vec{b}|^2 = 9 + 8 + 12 = 29 \] \[ |\vec{a} + \vec{b}| = \sqrt{29} \] Since $\vec{c}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$, the magnitude of the sum of all three vectors using the Pythagorean theorem is: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a} + \vec{b}|^2 + |\vec{c}|^2 \] \[ |\vec{a} + \vec{b} + \vec{c}|^2 = 29 + 25 = 54 \] \[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{54} = 3\sqrt{6} \]