Question

Define the points \(P(-4,1), Q(3,-4),\) and \(R(2,6) .\) Carry out the following calculations. Find two vectors parallel to \(\overrightarrow{R P}\) with length 4

Short answer

The two vectors parallel to RP with length 4 are (24/sqrt(61), 20/sqrt(61)) and (-24/sqrt(61), -20/sqrt(61)).

Step-by-step solution

Find the Vector RP

To find the vector RP, we subtract the coordinates of point P from the coordinates of point R. So, we have: RP = R - P RP = (2-(-4), 6-1) RP = (6, 5)

Find the Magnitude of RP

To find the magnitude of RP, or ||RP||, we use the Pythagorean theorem: ||RP|| = sqrt(6^2 + 5^2) ||RP|| = sqrt(36 + 25) ||RP|| = sqrt(61)

Find a Unit Vector in the Direction of RP

Now, we need to divide the vector RP by its magnitude to find a unit vector in the same direction. This will give us: u = (RP_x/||RP||, RP_y/||RP||) u = (6/sqrt(61), 5/sqrt(61))

Find Vectors Parallel to RP with Length 4

To find two vectors parallel to RP with length 4, we'll simply multiply the unit vector found in the previous step by 4 for one vector and -4 for the other vector. This gives us: v1 = 4u = (4(6/sqrt(61)), 4(5/sqrt(61))) v1 = (24/sqrt(61), 20/sqrt(61)) v2 = -4u = (-4(6/sqrt(61)), -4(5/sqrt(61))) v2 = (-24/sqrt(61), -20/sqrt(61)) So, the two vectors parallel to RP with length 4 are (24/sqrt(61), 20/sqrt(61)) and (-24/sqrt(61), -20/sqrt(61)).

Key concepts

Magnitude of a Vector

Understanding the magnitude of a vector is fundamental in vector calculations. The magnitude is essentially the length of the vector, representing the distance from its start point to its end point in space. To calculate it, we use the Pythagorean theorem:
This theorem states that for a vector \( \overrightarrow{RP} = (6, 5) \), the magnitude is calculated as follows:

• Calculate each component's square: \( 6^2 = 36 \) and \( 5^2 = 25 \).
• Add the squared components: \( 36 + 25 = 61 \).
• Take the square root of the sum to find the magnitude: \( \sqrt{61} \).

This gives us the length of \( \overrightarrow{RP} \), which is approximately 7.81. Knowing how to compute this helps in various fields, including physics and engineering.

Unit Vector

A unit vector is a vector with a magnitude of exactly 1. It maintains the direction of the original vector but reduces its size, making it perfect for finding direction. To convert any vector into a unit vector:

• Divide each component of the vector by its magnitude.

For a vector \( \overrightarrow{RP} = (6, 5) \) with a magnitude of \( \sqrt{61} \):

• Divide 6 by \( \sqrt{61} \) to get the x-component of the unit vector: \( \frac{6}{\sqrt{61}} \).
• Divide 5 by \( \sqrt{61} \) to get the y-component: \( \frac{5}{\sqrt{61}} \).

The resulting vector \( (\frac{6}{\sqrt{61}}, \frac{5}{\sqrt{61}}) \) points in the same direction as \( \overrightarrow{RP} \) but has a length of 1. Unit vectors are vital in simplifying complex vector equations and problems.

Parallel Vectors

Parallel vectors share the same or exact opposite direction without needing to have the same magnitude. For two vectors to be parallel, one must be a scalar multiple of the other. In other words, if you multiply one vector by a constant, you get a vector parallel to it.To find vectors parallel to \( \overrightarrow{RP} = (6, 5) \) with a specific length, say 4, follow these steps:

• Determine the unit vector of \( \overrightarrow{RP} \) as \( (\frac{6}{\sqrt{61}}, \frac{5}{\sqrt{61}}) \).
• Multiply by the desired length: here, 4 or -4 for the opposite direction.

This results in two parallel vectors: \((\frac{24}{\sqrt{61}}, \frac{20}{\sqrt{61}})\) in the same direction as \( \overrightarrow{RP} \) and \((-\frac{24}{\sqrt{61}}, -\frac{20}{\sqrt{61}})\), in the opposite direction.Understanding how vectors interact, particularly when they are parallel, is useful in fields such as physics, where forces need to be resolved into components.