Key Concepts and Formulas

• Angle between two vectors: The angle $\theta$ between two non-zero vectors $\vec{X}$ and $\vec{Y}$ is given by $\cos \theta = \frac{\vec{X} \cdot \vec{Y}}{|\vec{X}| |\vec{Y}|}$.
• Properties of perpendicular vectors: If $\vec{a}$ and $\vec{b}$ are perpendicular, then $\vec{a} \cdot \vec{b} = 0$.
• Magnitude of cross product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
• Properties of cross product with perpendicular vectors: If $\vec{a}$ and $\vec{b}$ are perpendicular, then $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}|$.
• Vector triple product identity (not directly used, but good to know): $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.
• Dot product distributivity: $\vec{X} \cdot (\vec{Y} + \vec{Z}) = \vec{X} \cdot \vec{Y} + \vec{X} \cdot \vec{Z}$.
• Scalar multiplication with dot product: $(k\vec{X}) \cdot \vec{Y} = k(\vec{X} \cdot \vec{Y})$.

Step-by-Step Solution

Step 1: Understand the Given Information and the Goal We are given two non-zero, perpendicular vectors $\vec{a}$ and $\vec{b}$ with equal magnitudes, i.e., $|\vec{a}| = |\vec{b}|$. We are also given the condition $|\vec{a} \times \vec{b}| = |\vec{a}|$. Our objective is to find the angle between the vector $\vec{V} = \vec{a} + \vec{b} + (\vec{a} \times \vec{b})$ and the vector $\vec{a}$.

Step 2: Simplify the Magnitude of the Cross Product Since $\vec{a}$ and $\vec{b}$ are perpendicular, the angle between them is $90^\circ$. Using the formula for the magnitude of the cross product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(90^\circ)$ $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| (1)$ We are given $|\vec{a}| = |\vec{b}|$. Let's denote this common magnitude by $k$. So, $|\vec{a}| = |\vec{b}| = k$. Then, $|\vec{a} \times \vec{b}| = k \cdot k = k^2$. We are also given $|\vec{a} \times \vec{b}| = |\vec{a}| = k$. Equating the two expressions for $|\vec{a} \times \vec{b}|$: $k^2 = k$ Since $\vec{a}$ is a non-zero vector, $k = |\vec{a}| \neq 0$. Thus, we can divide by $k$: $k = 1$. Therefore, we have $|\vec{a}| = |\vec{b}| = 1$ and $|\vec{a} \times \vec{b}| = 1$.

Step 3: Determine the Direction and Magnitude of the Cross Product Vector Let $\vec{c} = \vec{a} \times \vec{b}$. From Step 2, we know that $|\vec{c}| = |\vec{a} \times \vec{b}| = 1$. Also, since $\vec{a}$ and $\vec{b}$ are perpendicular, $\vec{c} = \vec{a} \times \vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$.

Step 4: Calculate the Dot Product of the Two Vectors We need to find the angle between $\vec{V} = \vec{a} + \vec{b} + \vec{c}$ and $\vec{a}$. Let this angle be $\theta$. Using the dot product formula: $\cos \theta = \frac{\vec{V} \cdot \vec{a}}{|\vec{V}| |\vec{a}|}$

First, let's compute the dot product $\vec{V} \cdot \vec{a}$: $\vec{V} \cdot \vec{a} = (\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}$ Using the distributive property of the dot product: $\vec{V} \cdot \vec{a} = (\vec{a} \cdot \vec{a}) + (\vec{b} \cdot \vec{a}) + (\vec{c} \cdot \vec{a})$

Now, let's evaluate each term:

• $\vec{a} \cdot \vec{a} = |\vec{a}|^2$. From Step 2, $|\vec{a}| = 1$, so $\vec{a} \cdot \vec{a} = 1^2 = 1$.
• $\vec{b} \cdot \vec{a} = 0$, because $\vec{a}$ and $\vec{b}$ are perpendicular.
• $\vec{c} \cdot \vec{a} = (\vec{a} \times \vec{b}) \cdot \vec{a}$. A vector resulting from a cross product is perpendicular to both of the original vectors. Therefore, $\vec{a} \times \vec{b}$ is perpendicular to $\vec{a}$. The dot product of two perpendicular vectors is zero. So, $\vec{c} \cdot \vec{a} = 0$.

Substituting these values back into the dot product equation: $\vec{V} \cdot \vec{a} = 1 + 0 + 0 = 1$.

Step 5: Calculate the Magnitude of the Vector $\vec{V}$ We need to find $|\vec{V}| = |\vec{a} + \vec{b} + \vec{c}|$. $|\vec{V}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c})$ $|\vec{V}|^2 = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c}$

Let's evaluate each term:

• $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1^2 = 1$.
• $\vec{a} \cdot \vec{b} = 0$ (perpendicular).
• $\vec{a} \cdot \vec{c} = \vec{a} \cdot (\vec{a} \times \vec{b}) = 0$ (perpendicular).
• $\vec{b} \cdot \vec{a} = 0$ (perpendicular).
• $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1^2 = 1$.
• $\vec{b} \cdot \vec{c} = \vec{b} \cdot (\vec{a} \times \vec{b})$. Since $\vec{c}$ is perpendicular to $\vec{b}$, this dot product is 0.
• $\vec{c} \cdot \vec{a} = (\vec{a} \times \vec{b}) \cdot \vec{a} = 0$ (perpendicular).
• $\vec{c} \cdot \vec{b} = (\vec{a} \times \vec{b}) \cdot \vec{b} = 0$ (perpendicular).
• $\vec{c} \cdot \vec{c} = |\vec{c}|^2$. From Step 2, $|\vec{c}| = 1$, so $\vec{c} \cdot \vec{c} = 1^2 = 1$.

Substituting these values: $|\vec{V}|^2 = 1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1 = 3$. So, $|\vec{V}| = \sqrt{3}$.

Step 6: Calculate the Cosine of the Angle Now we have all the components to find $\cos \theta$: $\cos \theta = \frac{\vec{V} \cdot \vec{a}}{|\vec{V}| |\vec{a}|}$ $\cos \theta = \frac{1}{(\sqrt{3})(1)}$ $\cos \theta = \frac{1}{\sqrt{3}}$

Step 7: Find the Angle The angle $\theta$ is given by $\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

Common Mistakes & Tips

• Confusing dot and cross products: Ensure you use the correct product for the operation (dot product for angle, cross product for perpendicularity and torque).
• Forgetting the properties of perpendicular vectors: The fact that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot (\vec{a} \times \vec{b}) = 0$ significantly simplifies calculations.
• Algebraic errors in dot product expansion: Carefully expand $|\vec{V}|^2$ and remember that the dot product is commutative ($\vec{X} \cdot \vec{Y} = \vec{Y} \cdot \vec{X}$).

Summary

We first used the given conditions, particularly the perpendicularity of $\vec{a}$ and $\vec{b}$ and the magnitude relation $|\vec{a} \times \vec{b}| = |\vec{a}|$, to deduce that $|\vec{a}| = |\vec{b}| = |\vec{a} \times \vec{b}| = 1$. Let $\vec{c} = \vec{a} \times \vec{b}$. We then calculated the dot product of the vector $\vec{a} + \vec{b} + \vec{c}$ with $\vec{a}$, which simplified to 1 due to the perpendicularity conditions. Subsequently, we calculated the magnitude of $\vec{a} + \vec{b} + \vec{c}$, which turned out to be $\sqrt{3}$. Finally, using the formula for the cosine of the angle between two vectors, we found $\cos \theta = \frac{1}{\sqrt{3}}$, leading to the angle $\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$.

The final answer is $\boxed{{\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)}$, which corresponds to option (D).