Problem 6 - Active Calculus Multivariable

Question

Let \(A=(5,0,0), B=(2,-2,-3),\) and \(P=(k, k, k) .\) The vector from \(A\) to \(B\) is perpendicular to the vector from \(A\) to \(P\) when \(k=\) ________

Step-by-Step Solution

Answer

k = \(\frac{15}{8}\)

Step 1: Finding the vectors AB and AP

We will first find the vector AB and the vector AP. The formula to find the vector AB is given by B - A, and for AP, it is given by P - A. Vector AB = B - A = (2 - 5, -2 - 0, -3 - 0) = (-3, -2, -3) Vector AP = P - A = (k - 5, k - 0, k - 0) = (k - 5, k, k)

Step 2: Finding the dot product and the condition for perpendicularity

Two vectors are said to be perpendicular if their dot product is equal to zero. We will find the dot product of vectors AB and AP and set it equal to zero to find the value of k. AB • AP = (-3, -2, -3) • (k - 5, k, k) = (-3)(k - 5) + (-2)k + (-3)k = 0

Step 3: Solving for k

Now we have an equation to solve for k: -3(k - 5) - 2k - 3k = 0 Expanding and gathering the terms we have: -3k + 15 - 2k - 3k = 0 -8k = -15 Dividing by -8: k = 15/8 The value of k for which the vector from A to B is perpendicular to the vector from A to P is k = 15/8.

Key Concepts

Dot Product and Perpendicular Vectors

When learning about vectors, a fundamental concept is the dot product—also known as the scalar product. It is a way to multiply two vectors, yielding a scalar (a single number, not a vector). The dot product is calculated by multiplying corresponding components of two vectors and then adding those products.

The formula for the dot product of two vectors \(\vec{a}\) and \(\vec{b}\), each with components \(a_i\) and \(b_i\), is:
\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]
Here's where it gets exciting: the dot product is directly related to the angle between the vectors. Specifically, if the dot product of two vectors is zero, we know that the vectors are perpendicular (or orthogonal) to each other. This is a crucial property used in geometry and physics, namely in our example where we want to find the value \(k\) that makes vector \(AB\) perpendicular to vector \(AP\).

Why is this helpful? If you're dealing with spatial problems, like finding vectors that outline a right angle, knowing how to calculate and interpret the dot product is indispensable.

Vector Subtraction

Vector subtraction is just as essential as addition. Understanding this operation allows us to find a vector that represents the direction and distance from one point to another. In our example, to find the vector \(AB\), we subtract the coordinates of point \(A\) from point \(B\), and similarly for vector \(AP\).

Here's the process step by step: \[\vec{AB} = \vec{B} - \vec{A}\] and \[\vec{AP} = \vec{P} - \vec{A}\]
The result of a vector subtraction is a new vector that points from one point to another. This is markedly useful in physics for displacement calculations or in computer graphics for motion algorithms.

Scalar Multiplication

Another basic vector operation is scalar multiplication, which involves multiplying a vector by a scalar (a real number), affecting its magnitude but not its direction, unless the scalar is negative, in which case the direction is reversed. For instance, when we solve for \(k\) in our problem, we're looking for a scalar that, when multiplied by the vector, will give us a resultant vector that is perpendicular to another.

Here’s how it works: if \(c\) is a scalar and \(\vec{v}\) is a vector, then \[c\vec{v} = (cv_1, cv_2, cv_3)\]
Scalar multiplication is pivotal in operations like stretching or compressing vectors, and changing a vector's length while keeping its orientation consistent. This comes in handy when dealing with physics problems involving force or velocity, as well as in 3D modeling to scale objects.