Key Concepts and Formulas

• Magnitude of Vector Sum and Difference: For any two vectors $\vec a$ and $\vec b$ with an angle $\theta$ between them, the magnitudes of their sum and difference are given by:
  • $||\vec a + \vec b||^2 = ||\vec a||^2 + ||\vec b||^2 + 2||\vec a|| ||\vec b|| \cos\theta$
  • $||\vec a - \vec b||^2 = ||\vec a||^2 + ||\vec b||^2 - 2||\vec a|| ||\vec b|| \cos\theta$
• Unit Vectors: A unit vector has a magnitude of 1. So, if $\vec a$ and $\vec b$ are unit vectors, $||\vec a|| = 1$ and $||\vec b|| = 1$.
• Trigonometric Half-Angle Identities: These are essential for simplifying expressions involving $1 \pm \cos\theta$:
  • $1 + \cos\theta = 2\cos^2(\theta/2)$
  • $1 - \cos\theta = 2\sin^2(\theta/2)$
• Maximization of Trigonometric Expressions: An expression of the form $A\cos x + B\sin x$ has a maximum value of $\sqrt{A^2 + B^2}$.

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Step-by-Step Solution

Let the given expression be $E$. We are asked to find the greatest value of $E = \sqrt{3} \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$.

Step 1: Simplify the magnitudes of the vector sum and difference using the properties of unit vectors.

• Reasoning: The problem involves the magnitudes of vector sums and differences, and we are given that $\vec a$ and $\vec b$ are unit vectors. Using the formulas for vector magnitudes and the fact that $||\vec a|| = ||\vec b|| = 1$ will allow us to express these magnitudes in terms of the angle $\theta$ between them.
• Let $\theta$ be the angle between the vectors $\vec a$ and $\vec b$, where $0 \le \theta \le \pi$. Since $\vec a$ and $\vec b$ are unit vectors, $||\vec a|| = 1$ and $||\vec b|| = 1$.
• For the magnitude of the sum: $||\vec a + \vec b||^2 = ||\vec a||^2 + ||\vec b||^2 + 2||\vec a|| ||\vec b|| \cos\theta$ Substituting $||\vec a|| = 1$ and $||\vec b|| = 1$: $||\vec a + \vec b||^2 = 1^2 + 1^2 + 2(1)(1)\cos\theta = 2 + 2\cos\theta = 2(1+\cos\theta)$ Taking the square root: $||\vec a + \vec b|| = \sqrt{2(1+\cos\theta)}$ Using the half-angle identity $1+\cos\theta = 2\cos^2(\theta/2)$: $||\vec a + \vec b|| = \sqrt{2(2\cos^2(\theta/2))} = \sqrt{4\cos^2(\theta/2)} = 2|\cos(\theta/2)|$ Since $0 \le \theta \le \pi$, we have $0 \le \theta/2 \le \pi/2$. In this interval, $\cos(\theta/2) \ge 0$, so $|\cos(\theta/2)| = \cos(\theta/2)$. Thus, $||\vec a + \vec b|| = 2\cos(\theta/2)$.
• For the magnitude of the difference: $||\vec a - \vec b||^2 = ||\vec a||^2 + ||\vec b||^2 - 2||\vec a|| ||\vec b|| \cos\theta$ Substituting $||\vec a|| = 1$ and $||\vec b|| = 1$: $||\vec a - \vec b||^2 = 1^2 + 1^2 - 2(1)(1)\cos\theta = 2 - 2\cos\theta = 2(1-\cos\theta)$ Taking the square root: $||\vec a - \vec b|| = \sqrt{2(1-\cos\theta)}$ Using the half-angle identity $1-\cos\theta = 2\sin^2(\theta/2)$: $||\vec a - \vec b|| = \sqrt{2(2\sin^2(\theta/2))} = \sqrt{4\sin^2(\theta/2)} = 2|\sin(\theta/2)|$ Since $0 \le \theta/2 \le \pi/2$, $\sin(\theta/2) \ge 0$, so $|\sin(\theta/2)| = \sin(\theta/2)$. Thus, $||\vec a - \vec b|| = 2\sin(\theta/2)$.

Step 2: Substitute the simplified magnitudes into the given expression.

• Reasoning: By substituting the expressions for $||\vec a + \vec b||$ and $||\vec a - \vec b||$ in terms of $\theta/2$, we transform the vector problem into a purely trigonometric one.
• Substitute $||\vec a + \vec b|| = 2\cos(\theta/2)$ and $||\vec a - \vec b|| = 2\sin(\theta/2)$ into the expression $E$: $E = \sqrt{3} (2\cos(\theta/2)) + (2\sin(\theta/2))$ $E = 2\sqrt{3}\cos(\theta/2) + 2\sin(\theta/2)$

Step 3: Find the maximum value of the resulting trigonometric expression.

• Reasoning: The expression for $E$ is now in the standard form $A\cos x + B\sin x$, where $x = \theta/2$. We can find its maximum value using the formula $\sqrt{A^2 + B^2}$.
• Let $x = \theta/2$. The expression is $E = 2\sqrt{3}\cos x + 2\sin x$.
• Here, $A = 2\sqrt{3}$ and $B = 2$.
• The maximum value of $E$ is given by $\sqrt{A^2 + B^2}$: $E_{\text{max}} = \sqrt{(2\sqrt{3})^2 + (2)^2}$ $E_{\text{max}} = \sqrt{(4 \times 3) + 4}$ $E_{\text{max}} = \sqrt{12 + 4}$ $E_{\text{max}} = \sqrt{16}$ $E_{\text{max}} = 4$
• To confirm this maximum is attainable, we can rewrite $E$ using the angle addition formula. Divide and multiply by $\sqrt{(2\sqrt{3})^2 + 2^2} = 4$: $E = 4 \left( \frac{2\sqrt{3}}{4}\cos(\theta/2) + \frac{2}{4}\sin(\theta/2) \right)$ $E = 4 \left( \frac{\sqrt{3}}{2}\cos(\theta/2) + \frac{1}{2}\sin(\theta/2) \right)$ Let $\cos\alpha = \frac{\sqrt{3}}{2}$ and $\sin\alpha = \frac{1}{2}$. Then $\alpha = \pi/6$. $E = 4 (\cos\alpha\cos(\theta/2) + \sin\alpha\sin(\theta/2))$ $E = 4 \cos(\theta/2 - \alpha)$ $E = 4 \cos(\theta/2 - \pi/6)$ Since $0 \le \theta \le \pi$, we have $0 \le \theta/2 \le \pi/2$. Thus, $-\pi/6 \le \theta/2 - \pi/6 \le \pi/2 - \pi/6 = \pi/3$. The maximum value of $\cos(\phi)$ for $\phi \in [-\pi/6, \pi/3]$ is 1, which occurs when $\phi = 0$. So, the maximum value of $E$ is $4 \times 1 = 4$.

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Common Mistakes & Tips

• Forgetting Absolute Values: When taking square roots of squared trigonometric functions (e.g., $\sqrt{\cos^2(\theta/2)}$), remember that $\sqrt{x^2} = |x|$. However, in this specific problem, the range of $\theta/2$ ($0 \le \theta/2 \le \pi/2$) ensures that $\cos(\theta/2)$ and $\sin(\theta/2)$ are non-negative, so the absolute values can be removed directly.
• Incorrect Trigonometric Identities: Ensure accurate application of half-angle identities ($1+\cos\theta$ and $1-\cos\theta$) and angle addition/subtraction formulas.
• Range of Angle: Pay close attention to the range of the angle $\theta$ and consequently $\theta/2$ when determining the maximum or minimum values of trigonometric functions.

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Summary

The problem asks for the greatest value of a vector expression involving the magnitudes of vector sums and differences of two unit vectors. By utilizing the formulas for vector magnitudes, the properties of unit vectors, and trigonometric half-angle identities, we transformed the expression into a trigonometric function of the angle between the vectors. The expression simplified to $2\sqrt{3}\cos(\theta/2) + 2\sin(\theta/2)$. This trigonometric expression was then maximized using the standard method for functions of the form $A\cos x + B\sin x$, yielding a maximum value of $\sqrt{(2\sqrt{3})^2 + 2^2} = \sqrt{12+4} = \sqrt{16} = 4$.

The final answer is $\boxed{3}$.