Question

The angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{5\pi}{6} \) and the projection of \( \vec{a} \) on \( \vec{b} \) is \( \frac{-9}{\sqrt{3}} \), then \( |\vec{a}| \) is equal to:

• A. $12$
• B. $8$
• C. $10$
• D. $4$
• E. $6$

Hint:

Remember that the projection is a scalar. A negative projection correctly indicates that the angle between the vectors is obtuse ($90^\circ < \theta \le 180^\circ$), which is consistent with our angle of $150^\circ$.

Answer 1

Answer: (E)

Concept: The scalar projection of vector $\vec{a}$ onto vector $\vec{b}$ is given by the formula $|\vec{a}| \cos \theta$, where $\theta$ is the angle between the two vectors. This value represents the magnitude of the component of $\vec{a}$ in the direction of $\vec{b}$, which can be negative if the angle is obtuse.

Step 1: Identify the given components from the problem.
From the image Screenshot 2026-05-01 023229.jpg:
• Angle $\theta = \frac{5\pi}{6}$.
• Scalar projection of $\vec{a}$ on $\vec{b} = \frac{-9}{\sqrt{3}}$.

Step 2: Set up the projection equation.
The formula for projection is: \[ \text{Proj}_{\vec{b}}\vec{a} = |\vec{a}| \cos \theta \] Substitute the given values into the equation: \[ |\vec{a}| \cos\left(\frac{5\pi}{6}\right) = \frac{-9}{\sqrt{3}} \]

Step 3: Calculate the value of the trigonometric function and solve for $|\vec{a}|$.
The angle $\frac{5\pi}{6}$ is in the second quadrant ($150^\circ$), where cosine is negative: \[ \cos\left(\frac{5\pi}{6}\right) = \cos\left(\pi - \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \] Now, substitute this back into the equation: \[ |\vec{a}| \left( -\frac{\sqrt{3}}{2} \right) = \frac{-9}{\sqrt{3}} \] Multiply both sides by $-1$ and rearrange to isolate $|\vec{a}|$: \[ |\vec{a}| = \frac{9}{\sqrt{3}} \times \frac{2}{\sqrt{3}} = \frac{18}{3} = 6 \]