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Math  /  Geometry

QuestionAssignment 3.2 and 4.1 (Question 8) Print
Name: Jude Elsayed Date: 2024-11-13 Official Time: 10:07:47 Question 8 [3 points] Find the vector equation for the line passing through the points P1(10,−2,2)P_{1}(10,-2,2) and P2(3,−4,3)P_{2}(3,-4,3). [xyz]=[000]+t[000]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]+t\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] SUBMIT AND MARK

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a line in 3D that goes through two given points. Watch out! Don't mix up the starting point and the direction vector.

STEP 2

1. Find a point on the line
2. Find the direction vector
3. Assemble the vector equation

STEP 3

We can use either of the given points as our starting point.
Let's choose P1P_1.
So, our **initial point** is (10,−2,2)(10, -2, 2).
This means we can write the first part of our vector equation as: [xyz]=[10−22]+t[]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ -2 \\ 2 \end{bmatrix} + t \begin{bmatrix} \\ \\ \end{bmatrix} We're almost there!

STEP 4

To find the **direction vector**, we subtract the coordinates of the **initial point**, P1P_1, from the coordinates of the other point, P2P_2.
Think of it like finding the "steps" to get from P1P_1 to P2P_2.

STEP 5

So, our calculation is: [3−43]−[10−22]=[−7−21]\begin{bmatrix} 3 \\ -4 \\ 3 \end{bmatrix} - \begin{bmatrix} 10 \\ -2 \\ 2 \end{bmatrix} = \begin{bmatrix} -7 \\ -2 \\ 1 \end{bmatrix} This gives us our **direction vector**: (−7,−2,1)(-7, -2, 1).
This tells us how much to change xx, yy, and zz as we move along the line.

STEP 6

Now, we just plug our **initial point** and **direction vector** into the vector equation format.

STEP 7

The vector equation of a line is given by: [xyz]=[x0y0z0]+t[abc]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix} + t \begin{bmatrix} a \\ b \\ c \end{bmatrix} where (x0,y0,z0)(x_0, y_0, z_0) is a **point on the line** and (a,b,c)(a, b, c) is the **direction vector**.

STEP 8

Putting it all together, we get: [xyz]=[10−22]+t[−7−21]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ -2 \\ 2 \end{bmatrix} + t \begin{bmatrix} -7 \\ -2 \\ 1 \end{bmatrix} Boom! We did it!

STEP 9

The vector equation of the line passing through P1(10,−2,2)P_1(10, -2, 2) and P2(3,−4,3)P_2(3, -4, 3) is: [xyz]=[10−22]+t[−7−21]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ -2 \\ 2 \end{bmatrix} + t \begin{bmatrix} -7 \\ -2 \\ 1 \end{bmatrix}

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