Key Concepts and Formulas

1. Dot Product Properties: For any vectors $\vec x, \vec y, \vec z$ and scalar $k$:

   • $\vec x \cdot \vec y = \vec y \cdot \vec x$ (Commutative)
   • $\vec x \cdot (\vec y + \vec z) = \vec x \cdot \vec y + \vec x \cdot \vec z$ (Distributive)
   • $|\vec v|^2 = \vec v \cdot \vec v$. For a unit vector, $|\vec v|=1$, so $\vec v \cdot \vec v = 1$.
   • $|\vec x + \vec y + \vec z|^2 = |\vec x|^2 + |\vec y|^2 + |\vec z|^2 + 2(\vec x \cdot \vec y + \vec y \cdot \vec z + \vec z \cdot \vec x)$.

2. Cross Product Properties: For any vectors $\vec x, \vec y, \vec z$:

   • $\vec x \times \vec x = \vec 0$ (Zero vector)
   • $\vec x \times \vec y = -(\vec y \times \vec x)$ (Anti-commutative)
   • $\vec x \times (\vec y + \vec z) = \vec x \times \vec y + \vec x \times \vec z$ (Distributive)
   • If $\vec x + \vec y + \vec z = \vec 0$, then $\vec x \times (\vec x + \vec y + \vec z) = \vec 0$, which simplifies to $\vec x \times \vec y + \vec x \times \vec z = \vec 0$.

Step-by-Step Solution

We are given three unit vectors $\overrightarrow a, \overrightarrow b, \overrightarrow c$ such that $|\overrightarrow a| = |\overrightarrow b| = |\overrightarrow c| = 1$ and $\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0$. We need to find the ordered pair $(\lambda, \overrightarrow d)$, where $\lambda = \overrightarrow a \cdot \vec b + \vec b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a$ and $\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a$.

Step 1: Calculate $\lambda$

To find $\lambda$, we will use the given condition $\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0$ and the properties of the dot product.

1. Square the vector sum: Start with the given equation: $\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0$. Take the dot product of this equation with itself: $(\overrightarrow a + \vec b + \overrightarrow c) \cdot (\overrightarrow a + \vec b + \overrightarrow c) = \overrightarrow 0 \cdot \overrightarrow 0$ The right side is $0$.

2. Expand the dot product: Using the identity $|\vec x + \vec y + \vec z|^2 = |\vec x|^2 + |\vec y|^2 + |\vec z|^2 + 2(\vec x \cdot \vec y + \vec y \cdot \vec z + \vec z \cdot \vec x)$: $|\overrightarrow a|^2 + |\overrightarrow b|^2 + |\overrightarrow c|^2 + 2(\overrightarrow a \cdot \vec b + \vec b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a) = 0$

3. Substitute known values: Since $\overrightarrow a, \overrightarrow b, \overrightarrow c$ are unit vectors, $|\overrightarrow a| = |\overrightarrow b| = |\overrightarrow c| = 1$. $1^2 + 1^2 + 1^2 + 2(\overrightarrow a \cdot \vec b + \vec b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a) = 0$ $1 + 1 + 1 + 2(\overrightarrow a \cdot \vec b + \vec b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a) = 0$ $3 + 2(\overrightarrow a \cdot \vec b + \vec b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a) = 0$

4. Use the definition of $\lambda$: We are given $\lambda = \overrightarrow a \cdot \vec b + \vec b \cdot \overrightarrow c + \overrightarrow c \cdot \overrightarrow a$. Substituting this into the equation: $3 + 2\lambda = 0$

5. Solve for $\lambda$: $2\lambda = -3$ $\lambda = -\frac{3}{2}$

Step 2: Calculate $\overrightarrow d$

To find $\overrightarrow d$, we will use the condition $\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0$ and the properties of the cross product.

1. Take cross product with $\overrightarrow a$: From $\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0$, take the cross product with $\overrightarrow a$: $\overrightarrow a \times (\overrightarrow a + \vec b + \overrightarrow c) = \overrightarrow a \times \overrightarrow 0$ Using the distributive property and $\overrightarrow a \times \overrightarrow a = \overrightarrow 0$: $\overrightarrow a \times \overrightarrow a + \overrightarrow a \times \vec b + \overrightarrow a \times \overrightarrow c = \overrightarrow 0$ $\overrightarrow 0 + \overrightarrow a \times \vec b + \overrightarrow a \times \overrightarrow c = \overrightarrow 0$ $\overrightarrow a \times \vec b = -\overrightarrow a \times \overrightarrow c$ Using the anti-commutative property $\overrightarrow x \times \overrightarrow y = -(\overrightarrow y \times \overrightarrow x)$, we have $-\overrightarrow a \times \overrightarrow c = \overrightarrow c \times \overrightarrow a$. Thus, we get: $\overrightarrow a \times \vec b = \overrightarrow c \times \overrightarrow a \quad (*)$

2. Take cross product with $\overrightarrow b$: Similarly, from $\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0$, take the cross product with $\overrightarrow b$: $\overrightarrow b \times (\overrightarrow a + \vec b + \overrightarrow c) = \overrightarrow b \times \overrightarrow 0$ Using the distributive property and $\overrightarrow b \times \vec b = \overrightarrow 0$: $\overrightarrow b \times \overrightarrow a + \overrightarrow b \times \vec b + \overrightarrow b \times \overrightarrow c = \overrightarrow 0$ $\overrightarrow b \times \overrightarrow a + \overrightarrow 0 + \overrightarrow b \times \overrightarrow c = \overrightarrow 0$ $\overrightarrow b \times \overrightarrow c = -\overrightarrow b \times \overrightarrow a$ Using the anti-commutative property, $-\overrightarrow b \times \overrightarrow a = \overrightarrow a \times \vec b$. Thus, we get: $\overrightarrow b \times \overrightarrow c = \overrightarrow a \times \vec b \quad (**)$

3. Combine the results: From equations $(*)$ and $(**)$, we have: $\overrightarrow a \times \vec b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a$

4. Substitute into the expression for $\overrightarrow d$: We are given $\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a$. Since all three terms are equal, we can substitute: $\overrightarrow d = (\overrightarrow a \times \vec b) + (\overrightarrow a \times \vec b) + (\overrightarrow a \times \vec b)$ $\overrightarrow d = 3(\overrightarrow a \times \vec b)$

Step 3: Form the Ordered Pair and Select the Correct Option

We have found $\lambda = -\frac{3}{2}$ and $\overrightarrow d = 3(\overrightarrow a \times \vec b)$. The ordered pair is $\left( {-\frac{3}{2}, 3(\overrightarrow a \times \vec b)} \right)$.

Now, let's compare this with the given options: (A) $\left( {{3 \over 2},3\overrightarrow a \times \overrightarrow c } \right)$ (B) $\left( { - {3 \over 2},3\overrightarrow c \times \overrightarrow b } \right)$ (C) $\left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right)$ (D) $\left( {{3 \over 2},3\overrightarrow b \times \overrightarrow c } \right)$

Our calculated $\lambda = -\frac{3}{2}$ matches options (B) and (C). Our calculated $\overrightarrow d = 3(\overrightarrow a \times \vec b)$ directly matches the vector component in option (C).

Therefore, the ordered pair is $\left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right)$.

Common Mistakes & Tips

• Dot Product Expansion: Remember the formula $|\vec x + \vec y + \vec z|^2 = |\vec x|^2 + |\vec y|^2 + |\vec z|^2 + 2(\vec x \cdot \vec y + \vec y \cdot \vec z + \vec z \cdot \vec x)$. Forgetting the factor of 2 is a common error.
• Cross Product Anti-commutativity: Be careful with signs when using $\vec x \times \vec y = -(\vec y \times \vec x)$. For instance, $-\overrightarrow a \times \overrightarrow c$ is equal to $\overrightarrow c \times \overrightarrow a$, not $-\overrightarrow c \times \overrightarrow a$.
• Geometric Interpretation: The condition $\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0$ for unit vectors implies that they form an equilateral triangle in space. The angle between any two of them is $120^\circ$. This can be used to verify the dot product result: $\overrightarrow a \cdot \vec b = |\overrightarrow a||\overrightarrow b|\cos(120^\circ) = 1 \cdot 1 \cdot (-\frac{1}{2}) = -\frac{1}{2}$. Summing these three yields $\lambda = 3 \times (-\frac{1}{2}) = -\frac{3}{2}$.

Summary

We utilized the properties of dot and cross products along with the given condition that three unit vectors sum to the zero vector. By squaring the vector sum, we found the scalar component $\lambda = -\frac{3}{2}$. By taking cross products of the sum with individual vectors, we established the relationship $\overrightarrow a \times \vec b = \overrightarrow b \times \overrightarrow c = \overrightarrow c \times \overrightarrow a$, which allowed us to simplify the vector component $\overrightarrow d$ to $3(\overrightarrow a \times \vec b)$. The resulting ordered pair is $\left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right)$.

The final answer is $\boxed{\left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right)}$ which corresponds to option (C).