Question

If $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} + \vec{b}$ makes an angle of $60^\circ$ with $\vec{b}$, then $|\vec{a}|$ is equal to:

• A. $0$
• B. $\frac{1}{\sqrt{3}} |\vec{b}|$
• C. $\frac{1}{|\vec{b}|}$
• D. $|\vec{b}|$
• E. $\sqrt{3} |\vec{b}|$

Hint:

For perpendicular vectors $\vec{a}$ and $\vec{b}$, the vector $\vec{a} + \vec{b}$ forms a right triangle. The angle it makes with $\vec{b}$ can be solved simply using $\tan \theta = \frac{|\vec{a}|}{|\vec{b}|}$. Here $\tan 60^\circ = \sqrt{3}$, so $|\vec{a}|/|\vec{b}| = \sqrt{3}$.

Answer 1

Answer: (E)

Concept: The angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$. Here, $\vec{a} \cdot \vec{b} = 0$ means $\vec{a}$ and $\vec{b}$ are perpendicular.

Step 1: Use the angle formula for $\vec{a} + \vec{b}$ and $\vec{b}$.
Let $\vec{u} = \vec{a} + \vec{b}$ and $\vec{v} = \vec{b}$. The angle between them is $60^\circ$: \[ \cos 60^\circ = \frac{(\vec{a} + \vec{b}) \cdot \vec{b}}{|\vec{a} + \vec{b}| |\vec{b}|} \]

Step 2: Simplify the dot product and magnitude.
$(\vec{a} + \vec{b}) \cdot \vec{b} = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$. Since $\vec{a} \cdot \vec{b} = 0$, this equals $|\vec{b}|^2$. Since $\vec{a} \perp \vec{b}$, by the Pythagorean theorem for vectors: $|\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2}$.

Step 3: Solve for $|\vec{a}|$.
\[ \frac{1}{2} = \frac{|\vec{b}|^2}{\sqrt{|\vec{a}|^2 + |\vec{b}|^2} |\vec{b}|} \quad \Rightarrow \quad \frac{1}{2} = \frac{|\vec{b}|}{\sqrt{|\vec{a}|^2 + |\vec{b}|^2}} \] Squaring both sides: \[ \frac{1}{4} = \frac{|\vec{b}|^2}{|\vec{a}|^2 + |\vec{b}|^2} \quad \Rightarrow \quad |\vec{a}|^2 + |\vec{b}|^2 = 4|\vec{b}|^2 \] \[ |\vec{a}|^2 = 3|\vec{b}|^2 \quad \Rightarrow \quad |\vec{a}| = \sqrt{3} |\vec{b}| \]