Questions: Show that for any two nonzero vectors A and B in any number D of euclidean spatial dimensions, the magnitude of their sum C=A+B is maximum when the two vectors are parallel and minimum when they are anti-parallel.

Show that for any two nonzero vectors A and B in any number D of euclidean spatial dimensions, the magnitude of their sum C=A+B is maximum when the two vectors are parallel and minimum when they are anti-parallel.

Transcript text: Show that for any two nonzero vectors $\vec{A}$ and $\vec{B}$ in any number $D$ of euclidean spatial dimensions, the magnitude of their sum $\vec{C}=\vec{A}+\vec{B}$ is maximum when the two vectors are parallel and minimum when they are anti-parallel.

Solution

Solution Steps

Step 1: Define the Magnitude of the Sum of Two Vectors

Given two nonzero vectors $\vec{A}$ and $\vec{B}$, we need to find the magnitude of their sum $\vec{C} = \vec{A} + \vec{B}$. The magnitude of $\vec{C}$ is given by: \[ |\vec{C}| = |\vec{A} + \vec{B}| \]

Step 2: Use the Magnitude Formula

The magnitude of a vector $\vec{C}$ can be expressed using the dot product: \[ |\vec{C}| = \sqrt{(\vec{A} + \vec{B}) \cdot (\vec{A} + \vec{B})} \] Expanding the dot product, we get: \[ |\vec{C}| = \sqrt{\vec{A} \cdot \vec{A} + 2 \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{B}} \] Since $\vec{A} \cdot \vec{A} = |\vec{A}|^2$ and $\vec{B} \cdot \vec{B} = |\vec{B}|^2$, we can rewrite this as: \[ |\vec{C}| = \sqrt{|\vec{A}|^2 + 2 \vec{A} \cdot \vec{B} + |\vec{B}|^2} \]

Step 3: Express the Dot Product in Terms of Magnitudes and Angle

The dot product $\vec{A} \cdot \vec{B}$ can be expressed in terms of the magnitudes of $\vec{A}$ and $\vec{B}$ and the cosine of the angle $\theta$ between them: \[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \] Substituting this into the magnitude formula, we get: \[ |\vec{C}| = \sqrt{|\vec{A}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta + |\vec{B}|^2} \]

Step 4: Analyze the Conditions for Maximum and Minimum Magnitude

To find the conditions for maximum and minimum magnitude, we need to consider the value of $\cos \theta$:

When $\vec{A}$ and $\vec{B}$ are parallel, $\theta = 0$ and $\cos \theta = 1$: \[ |\vec{C}| = \sqrt{|\vec{A}|^2 + 2 |\vec{A}| |\vec{B}| + |\vec{B}|^2} = \sqrt{(|\vec{A}| + |\vec{B}|)^2} = |\vec{A}| + |\vec{B}| \]
When $\vec{A}$ and $\vec{B}$ are anti-parallel, $\theta = \pi$ and $\cos \theta = -1$: \[ |\vec{C}| = \sqrt{|\vec{A}|^2 - 2 |\vec{A}| |\vec{B}| + |\vec{B}|^2} = \sqrt{(|\vec{A}| - |\vec{B}|)^2} = ||\vec{A}| - |\vec{B}|| \]