Question

the three vectors shown are dimensionless (their lengths are pure numbers with no units), and satisfy the equation: 2\vec{a} + \vec{b} = \vec{c}
furthermore, \vec{b} and \vec{c} are perpendicular to each other, and the angle \theta is restricted to the range 0 \leq \theta \leq 90^\circ.
the acute angle formed between \vec{a} and \vec{c} is 15.2 degrees. find the magnitude of \vec{a}. provide at least one decimal place

Explanation:

Step1: Identify vector $\vec{B}$

From the grid, $\vec{B} = -4\hat{i} + 6\hat{j}$, so $|\vec{B}| = \sqrt{(-4)^2 + 6^2} = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$

Step2: Rearrange given vector equation

$\vec{B} = \vec{C} - 2\vec{A}$

Step3: Use perpendicularity condition

Since $\vec{B} \perp \vec{C}$, their dot product is 0:
$\vec{B} \cdot \vec{C} = 0$
Substitute $\vec{B}$:
$(\vec{C} - 2\vec{A}) \cdot \vec{C} = 0$
$\vec{C} \cdot \vec{C} - 2\vec{A} \cdot \vec{C} = 0$
$|\vec{C}|^2 = 2|\vec{A}||\vec{C}|\cos(15.2^\circ)$
Cancel $|\vec{C}|$ (non-zero):
$|\vec{C}| = 2|\vec{A}|\cos(15.2^\circ)$

Step4: Take magnitude of $\vec{B} = \vec{C} - 2\vec{A}$

Square both sides:
$|\vec{B}|^2 = |\vec{C}|^2 + 4|\vec{A}|^2 - 4\vec{A} \cdot \vec{C}$
Substitute $\vec{A} \cdot \vec{C} = |\vec{A}||\vec{C}|\cos(15.2^\circ)$ and $|\vec{C}| = 2|\vec{A}|\cos(15.2^\circ)$:
$|\vec{B}|^2 = (2|\vec{A}|\cos15.2^\circ)^2 + 4|\vec{A}|^2 - 4|\vec{A}|(2|\vec{A}|\cos15.2^\circ)\cos15.2^\circ$
Simplify:
$|\vec{B}|^2 = 4|\vec{A}|^2\cos^215.2^\circ + 4|\vec{A}|^2 - 8|\vec{A}|^2\cos^215.2^\circ$
$|\vec{B}|^2 = 4|\vec{A}|^2 - 4|\vec{A}|^2\cos^215.2^\circ$
$|\vec{B}|^2 = 4|\vec{A}|^2(1-\cos^215.2^\circ)$
Use $1-\cos^2\theta = \sin^2\theta$:
$|\vec{B}|^2 = 4|\vec{A}|^2\sin^215.2^\circ$

Step5: Solve for $|\vec{A}|$

$|\vec{A}| = \frac{|\vec{B}|}{2\sin(15.2^\circ)}$
Substitute $|\vec{B}|=\sqrt{52}$, $\sin(15.2^\circ)\approx0.2623$:
$|\vec{A}| = \frac{\sqrt{52}}{2\times0.2623} \approx \frac{7.211}{0.5246} \approx 13.7$

Answer:

$\approx 13.7$