Question

If $|\vec{a}| = 2$, $|\vec{b}| = 3$, $|\vec{c}| = 5$ and each of the angles between the vectors $\vec{a}$ and $\vec{b}$, $\vec{b}$ and $\vec{c}$, $\vec{c}$ and $\vec{a}$ is $60^\circ$, then the value of $|\vec{a} + \vec{b} + \vec{c}|$ is

• A. $\sqrt{69}$
• B. $\sqrt{70}$
• C. $\sqrt{80}$
• D. $\sqrt{39}$

Hint:

For the sum of three vectors, remember the expansion formula: $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$. Also recall $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$.

Answer 1

The Correct Option is A

Step 1: Calculate the dot products between each pair of vectors. The dot product of two vectors $\vec{u}$ and $\vec{v}$ is given by $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$, where $\theta$ is the angle between them. We are given $|\vec{a}| = 2$, $|\vec{b}| = 3$, $|\vec{c}| = 5$, and the angle between any pair of vectors is $60^\circ$. For $\vec{a} \cdot \vec{b}$: \[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(60^\circ) = (2)(3)\left(\frac{1}{2}\right) = 3 \] For $\vec{b} \cdot \vec{c}$: \[ \vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos(60^\circ) = (3)(5)\left(\frac{1}{2}\right) = \frac{15}{2} \] For $\vec{c} \cdot \vec{a}$: \[ \vec{c} \cdot \vec{a} = |\vec{c}||\vec{a}|\cos(60^\circ) = (5)(2)\left(\frac{1}{2}\right) = 5 \]
Step 2: Use the formula for the magnitude of the sum of three vectors. The square of the magnitude of the sum of three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is given by: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
Step 3: Substitute the known values into the formula. We have $|\vec{a}| = 2$, $|\vec{b}| = 3$, $|\vec{c}| = 5$, $\vec{a} \cdot \vec{b} = 3$, $\vec{b} \cdot \vec{c} = \frac{15}{2}$, and $\vec{c} \cdot \vec{a} = 5$. \[ |\vec{a} + \vec{b} + \vec{c}|^2 = (2)^2 + (3)^2 + (5)^2 + 2\left(3 + \frac{15}{2} + 5\right) \] \[ = 4 + 9 + 25 + 2\left(8 + \frac{15}{2}\right) \] \[ = 38 + 2\left(\frac{16}{2} + \frac{15}{2}\right) \] \[ = 38 + 2\left(\frac{31}{2}\right) \] \[ = 38 + 31 \] \[ = 69 \]
Step 4: Find the magnitude of $\vec{a + \vec{b} + \vec{c}$.} \[ |\vec{a} + \vec{b} + \vec{c}| = \sqrt{69} \]
Step 5: State the final answer. The value of $|\vec{a} + \vec{b} + \vec{c}|$ is $\sqrt{69}$.