To solve for the dot product of vectors $\vec{a}$ and $\vec{b}$, we can use the formula:

$\vec{a} \cdot \vec{b} = a_x \cdot b_x + a_y \cdot b_y + a_z \cdot b_z$

where:

• $\vec{a} = -\hat{i} + 2\hat{j} - 3\hat{k}$,
• $\vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$.

Step 1: Identify the components

From the vectors, we have:

• $a_x = -1$, $a_y = 2$, $a_z = -3$,
• $b_x = 2$, $b_y = -3$, $b_z = 5$.

Step 2: Calculate each component of the dot product

$a_x \cdot b_x = (-1) \cdot 2 = -2$ $a_y \cdot b_y = 2 \cdot (-3) = -6$ $a_z \cdot b_z = (-3) \cdot 5 = -15$

Step 3: Sum the results

$\vec{a} \cdot \vec{b} = -2 - 6 - 15 = -23$

Final Answer

$\vec{a} \cdot \vec{b} = -23$

Would you like more details or have any questions?

Related Questions

1. What is the geometric interpretation of the dot product result being negative?
2. How do you find the angle between two vectors using their dot product?
3. What would the cross product $\vec{a} \times \vec{b}$ look like?
4. How does changing the magnitude of a vector affect its dot product with another vector?
5. How does the dot product help determine if two vectors are perpendicular?

Tip: The dot product of two vectors is zero if they are perpendicular, since $\cos(90^\circ) = 0$.