© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Express the square of the magnitude of the sum of vectors
Since the vectors \overrightarrow{a} , \overrightarrow{b} , and \overrightarrow{c} are mutually perpendicular and of the same magnitude, let |\overrightarrow{a}| = |\overrightarrow{b}| = |\overrightarrow{c}| = 1 for simplicity. The square of the magnitude of \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} is expanded as:
|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|^2 = |\overrightarrow{a}|^2 + |\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 \;+\; 2 \bigl(\overrightarrow{a} \cdot \overrightarrow{b} \;+\; \overrightarrow{a} \cdot \overrightarrow{c} \;+\; \overrightarrow{b} \cdot \overrightarrow{c}\bigr).
Step 2: Use orthogonality to simplify the dot products
Because \overrightarrow{a} , \overrightarrow{b} , and \overrightarrow{c} are mutually perpendicular:
\overrightarrow{a} \cdot \overrightarrow{b} = 0,\quad \overrightarrow{b} \cdot \overrightarrow{c} = 0,\quad \overrightarrow{c} \cdot \overrightarrow{a} = 0.
Also, each has magnitude 1, so |\overrightarrow{a}|^2 = |\overrightarrow{b}|^2 = |\overrightarrow{c}|^2 = 1. Hence,
|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|^2 = 1 + 1 + 1 + 2(0 + 0 + 0) = 3.
Therefore,
|\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}| = \sqrt{3}.
Step 3: Relate the angle \theta to the dot product
Each vector \overrightarrow{a} , \overrightarrow{b} , and \overrightarrow{c} is equally inclined at an angle \theta with the vector \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} . Focusing on \overrightarrow{a} , we use the dot product:
\overrightarrow{a} \cdot (\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}) = |\overrightarrow{a}|\; |\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}|\; \cos \theta.
Since |\overrightarrow{a}| = 1 , the left-hand side simplifies to:
\overrightarrow{a} \cdot \overrightarrow{a} + \overrightarrow{a} \cdot \overrightarrow{b} + \overrightarrow{a} \cdot \overrightarrow{c} = 1 + 0 + 0 = 1.
On the right-hand side, we get:
1 = (\sqrt{3}) \cos \theta.
Thus,
\cos \theta = \frac{1}{\sqrt{3}}.
Step 4: Find \cos 2\theta and compute the required expression
Using the double-angle identity for cosine:
\cos 2\theta = 2 \cos^2 \theta - 1.
Substitute \cos \theta = \frac{1}{\sqrt{3}} :
\cos 2\theta = 2 \left(\frac{1}{\sqrt{3}}\right)^2 - 1 = 2 \left(\frac{1}{3}\right) - 1 = \frac{2}{3} - 1 = -\frac{1}{3}.
Finally, compute 36 \cos^2(2\theta) :
36 \cos^2(2\theta) = 36 \left(-\frac{1}{3}\right)^2 = 36 \left(\frac{1}{9}\right) = 4.
Therefore, the value of 36 \cos^2(2\theta) is \boxed{4} .
