Question

Let $|\vec{a}|=2,|\vec{b}|=\sqrt{13+6\sqrt{3}}$ and $|\vec{c}|=3$. If $\vec{a}-\vec{b}+\vec{c}=0$, then the angle between $\vec{a}$ and $\vec{c}$ is

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{12}$
• D. $\frac{\pi}{2}$
• E. $\frac{\pi}{6}$

Hint:

Math Tip: Whenever an equation sums to zero (like $\vec{a}+\vec{b}+\vec{c}=0$) and you need the angle between two specific vectors, always isolate the third vector on the opposite side of the equation and square both sides.

Answer 1

Answer: (E)

Concept:
Vectors - Vector Algebra and Dot Product.
The magnitude of a sum of vectors satisfies $|\vec{u}+\vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2|\vec{u}||\vec{v}|\cos\theta$, where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$.
Step 1: Rearrange the given vector equation.
We are given the relation $\vec{a} - \vec{b} + \vec{c} = 0$.
Isolate the vector $\vec{b}$ on one side to group the vectors we want to find the angle between ($\vec{a}$ and $\vec{c}$): $$ \vec{a} + \vec{c} = \vec{b} $$
Step 2: Square both sides to utilize magnitudes.
Take the dot product of each side with itself (which squares the magnitudes): $$ |\vec{a} + \vec{c}|^2 = |\vec{b}|^2 $$
Step 3: Expand the squared binomial of vectors.
Using the property of dot products: $$ |\vec{a}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{c}) = |\vec{b}|^2 $$ Substitute the definition of the dot product ($\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos\theta$): $$ |\vec{a}|^2 + |\vec{c}|^2 + 2|\vec{a}||\vec{c}|\cos\theta = |\vec{b}|^2 $$
Step 4: Substitute the given numerical values.
We are given $|\vec{a}|=2$, $|\vec{c}|=3$, and $|\vec{b}|=\sqrt{13+6\sqrt{3}}$. $$ (2)^2 + (3)^2 + 2(2)(3)\cos\theta = \left(\sqrt{13+6\sqrt{3}}\right)^2 $$ $$ 4 + 9 + 12\cos\theta = 13 + 6\sqrt{3} $$
Step 5: Solve for $\cos\theta$.
Simplify the equation: $$ 13 + 12\cos\theta = 13 + 6\sqrt{3} $$ Subtract 13 from both sides: $$ 12\cos\theta = 6\sqrt{3} $$ $$ \cos\theta = \frac{6\sqrt{3}}{12} = \frac{\sqrt{3}}{2} $$
Step 6: Determine the angle $\theta$.
Since $\cos\theta = \frac{\sqrt{3}}{2}$, the angle $\theta$ in the principal range $[0, \pi]$ is: $$ \theta = \frac{\pi}{6} $$