This question involves a dot product and a cross product in vector algebra. Let's break it down step by step.

The expression given is:

$\vec{i} \cdot (\vec{j} \times \vec{k})$

Step 1: Recall the cross product of standard basis vectors

In 3D vector space, the standard unit vectors $\vec{i}$, $\vec{j}$, and $\vec{k}$ satisfy the following cross product rules:

• $\vec{j} \times \vec{k} = \vec{i}$
• $\vec{k} \times \vec{i} = \vec{j}$
• $\vec{i} \times \vec{j} = \vec{k}$

Thus, $\vec{j} \times \vec{k} = \vec{i}$.

Step 2: Substitute the result of the cross product

The expression becomes: $\vec{i} \cdot (\vec{j} \times \vec{k}) = \vec{i} \cdot \vec{i}$

Step 3: Compute the dot product

The dot product of a vector with itself is equal to the square of its magnitude. Since $\vec{i}$ is a unit vector: $\vec{i} \cdot \vec{i} = |\vec{i}|^2 = 1$

Final Answer:

$\boxed{1}$

Let me know if you'd like a detailed explanation or have further questions! Below are some related questions to deepen understanding.

5 Related Questions:

1. What is the result of $\vec{i} \cdot (\vec{i} \times \vec{j})$?
2. What is the magnitude of the cross product $\vec{a} \times \vec{b}$ in terms of $|\vec{a}|$, $|\vec{b}|$, and the angle between them?
3. Why is the dot product of orthogonal vectors zero?
4. What are the geometric interpretations of the dot product and cross product?
5. How are the properties of $\vec{i}$, $\vec{j}$, and $\vec{k}$ defined in higher dimensions?

Tip:

Always remember the cyclic nature of $\vec{i}$, $\vec{j}$, and $\vec{k}$ in the cross product: $\vec{i} \times \vec{j} = \vec{k}$, $\vec{j} \times \vec{k} = \vec{i}$, and $\vec{k} \times \vec{i} = \vec{j}$.