Problem 11 - Calculus: Early Transcendentals

Question

Let \(P=(4,5,6), Q=(1,2,-5) .\) Find \(\overrightarrow{P Q}\). Find a vector with the same direction as \(\overrightarrow{P Q}\) but with length 1 . Find a vector with the same direction as \(\overrightarrow{P Q}\) but with length \(4 .\)

Step-by-Step Solution

Answer

\(\overrightarrow{PQ} = (-3, -3, -11)\), unit vector: \(\left(\frac{-3}{\sqrt{139}}, \frac{-3}{\sqrt{139}}, \frac{-11}{\sqrt{139}}\right)\), length 4 vector: \(\left(\frac{-12}{\sqrt{139}}, \frac{-12}{\sqrt{139}}, \frac{-44}{\sqrt{139}}\right)\).

Step 1: Calculate Vector \(\overrightarrow{PQ}\)

To find \(\overrightarrow{PQ}\), subtract the coordinates of point \(P\) from point \(Q\): \(\overrightarrow{PQ} = Q - P \). Calculate each component separately: \[ \overrightarrow{PQ} = (1-4, 2-5, -5-6) = (-3, -3, -11) \].

Step 2: Find the Magnitude of \(\overrightarrow{PQ}\)

The magnitude (length) of vector \(\overrightarrow{PQ}\) is calculated using the formula: \(\|\overrightarrow{PQ}\| = \sqrt{(-3)^2 + (-3)^2 + (-11)^2}\). Simplify to find \[ \|\overrightarrow{PQ}\| = \sqrt{9 + 9 + 121} = \sqrt{139} \].

Step 3: Obtain a Unit Vector in the Direction of \(\overrightarrow{PQ}\)

A unit vector in the direction of \(\overrightarrow{PQ}\) is found by dividing each component of \(\overrightarrow{PQ}\) by its magnitude \( \|\overrightarrow{PQ}\| \). Thus, the unit vector is \[ \left(\frac{-3}{\sqrt{139}}, \frac{-3}{\sqrt{139}}, \frac{-11}{\sqrt{139}}\right) \].

Step 4: Find a Vector with Length 4 in the Direction of \(\overrightarrow{PQ}\)

To find a vector with the length 4 in the direction of \(\overrightarrow{PQ}\), multiply the unit vector by 4. The resulting vector is \[ \left(4 \times \frac{-3}{\sqrt{139}}, 4 \times \frac{-3}{\sqrt{139}}, 4 \times \frac{-11}{\sqrt{139}}\right) \].This simplifies to \( \left(\frac{-12}{\sqrt{139}}, \frac{-12}{\sqrt{139}}, \frac{-44}{\sqrt{139}}\right) \).

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude, or length, of exactly 1. It is a very useful tool in vector mathematics because it allows us to express the direction of any vector without altering its original path. The formula to find a unit vector is relatively simple. You just take each component of the vector you are looking at and divide it by the magnitude of the vector.
For example, if you have a vector \(\overrightarrow{PQ} = (-3, -3, -11)\), you first find the magnitude \(\|\overrightarrow{PQ},\| = \sqrt{139},\) as calculated earlier. Then, by dividing each component by this magnitude, you get the unit vector \(\left(\frac{-3}{\sqrt{139}}, \frac{-3}{\sqrt{139}}, \frac{-11}{\sqrt{139}}\right)\). The resultant unit vector maintains the original direction of \(\overrightarrow{PQ}\) but has a length of 1.

Vector Magnitude

The magnitude of a vector is its length in the geometric space. To find a vector's magnitude, we use the Pythagorean theorem extended into three dimensions.
If the vector is expressed in the form \((x, y, z)\), we calculate its magnitude using the formula \(\|v\|=\sqrt{x^2 + y^2 + z^2}\). This measure helps us understand the 'size' of the vector irrespective of its direction. In our exercise, the vector \(\overrightarrow{PQ} = (-3, -3, -11)\) has a magnitude \(\sqrt{139}\) determined by substituting the values into the formula.
This allows for determining not only the length of \(\overrightarrow{PQ}\) but also serves as the foundation for calculating a unit vector or scaling the vector to a desired length, as illustrated when creating a vector that is four times the unit vector.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometric figures through coordinate planes and the algebraic equations associated with them. It enables us to systematically handle geometric problems by translating them into an algebraic form.
In the context of vectors, coordinate geometry helps us understand and compute vector operations using coordinates of the points in the space. Given points \(P\) and \(Q\) in this exercise, we derive \(\overrightarrow{PQ}\) simply by performing arithmetic operations on their coordinates.
This approach empowers solving complex geometric problems by leveraging algebra, delivering results such as vector magnitudes or finding vectors with desired properties like a specific direction or length. Its blend of algebra and geometry makes it a powerful ally in solving various problems not only in mathematics but also in physics and engineering.