91Ó°ÊÓ

The dot product is a fundamental operation in vector algebra. It gives us a way to multiply two vectors and results in a scalar value. To compute the dot product, you simply multiply corresponding components of two vectors and then sum up the products.

Consider two vectors \( \mathbf{a} = [a_1, a_2, a_3] \) and \( \mathbf{b} = [b_1, b_2, b_3] \). The dot product, denoted as \( \mathbf{a} \cdot \mathbf{b} \), is calculated as:

\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

For our exercise, the vectors \( \mathbf{u} = [1, -1, 2] \) and \( \mathbf{v} = [k^2, k, -3] \) are given. By substituting these into the formula, we build the expression:

\( 1 \cdot k^2 + (-1) \cdot k + 2 \cdot (-3) = k^2 - k - 6 \).

The dot product gives crucial information about the relationship between two vectors. If the dot product is zero, the vectors are orthogonal, meaning they form a 90-degree angle with each other.

A quadratic equation is a second-order polynomial equation which can be expressed in the standard form: \( ax^2 + bx + c = 0 \).

In this form, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). Solving quadratic equations is vital in many areas, from physics to finance.

The general approach to solving these equations is through the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our problem, we had the equation \( k^2 - k - 6 = 0 \). Here, \( a = 1 \), \( b = -1 \), and \( c = -6 \). Applying the quadratic formula helps us find the possible values of \( k \) that make the vectors orthogonal.

Simplifying, we find:

• \( k = \frac{1 \pm \sqrt{25}}{2} \)
• Thus, \( k = 3 \) or \( k = -2 \).

This solution tells us for which scalar values the vectors intersect orthogonally.

A scalar is a simple quantity that only has magnitude, no direction. It's often used to scale or modify vector quantities. When we talk about the dot product as generating a scalar value, it means that the result of this product is just a number, not another vector.

In our context, the scalar \( k \) is the variable we're solving for, which influences the direction of vector \( \mathbf{v} \).

Scalars play a crucial role in adjusting vectors within mathematical operations, such as:

• Modifying the length or magnitude of a vector.
• Determining the result of vector algebra operations like the dot product.

By finding appropriate scalar values (here, \( k = 3 \) and \( k = -2 \)), we ensure that the two vectors are orthogonal. This concept allows vectors to be adapted and manipulated effectively in multidimensional spaces.