Find the vector and parametric equations for the line through the point P=(5,−2,3) and the point Q=(2,−7,8).

tinfoQ

tinfoQ

Answered question

2020-10-18

Find the vector and parametric equations for the line through the point P=(5,−2,3) and the point Q=(2,−7,8).

Answer & Explanation

comentezq

comentezq

Skilled2020-10-19Added 106 answers

Here think about your starting point, say it is P.
Start at (5,−2,3)(5,−2,3) then to get to Q, you must travel in the direction of Q which is given by Q−P=(2,−7,8)−(5,−2,3)=(−3,−5,5). Now, traveling this direction in any scaled value gives you a line determined by the following vector equation:
(5,−2,3)+t(−3,−5,5)
which you can write in parametric form:
x=5−3t, y=−2−3t, and z=3+5t
notice that all we did was just write it in a form that represents a formula for each point (x,y,z)(x,y,z).

RizerMix

RizerMix

Expert2023-05-25Added 656 answers

Step 1. Finding the Direction Vector (𝐯):
The direction vector 𝐯 is obtained by subtracting the coordinates of P from those of Q. Hence,
𝐯=[257(2)83]=[355]
Step 2. Parametric Equations:
Let x, y, and z represent the coordinates of any point on the line. We can express the line in parametric form as:
x=x0+aty=y0+btz=z0+ct
where (x0,y0,z0) represents the coordinates of a known point on the line (in this case, P), and a, b, and c are scalar parameters.
Substituting the known values, we have:
x=53ty=25tz=3+5t
Thus, the parametric equations of the line passing through P and Q are:
x=53ty=25tz=3+5t
Don Sumner

Don Sumner

Skilled2023-05-25Added 184 answers

To find the vector and parametric equations for the line passing through point P=(5,2,3) and point Q=(2,7,8), we can follow these steps:
1. Find the direction vector 𝐯 of the line by subtracting the coordinates of P from the coordinates of Q.
𝐯=[257(2)83]=[355]
2. Write the vector equation of the line using the point P and the direction vector 𝐯.
𝐫=[xyz]=[523]+t[355]
3. Write the parametric equations of the line by expanding the vector equation.
x=53t
y=25t
z=3+5t
Therefore, the vector equation of the line is 𝐫=[523]+t[355], and the parametric equations of the line are x=53t, y=25t, and z=3+5t.
Vasquez

Vasquez

Expert2023-05-25Added 669 answers

Result:
x(t)=53t, y(t)=25t, and z(t)=3+5t
Solution:
First, let's find the direction vector of the line. This can be done by subtracting the coordinates of the two points:
𝐯=𝐐𝐏=(2,7,8)(5,2,3)=(3,5,5).
The vector 𝐯 represents the direction of the line.
Now, let's write the parametric equations for the line. We'll use the point P as the starting point, and we'll introduce a parameter t to represent any point on the line:
𝐫(t)=𝐏+t𝐯.
Substituting the values, we have:
𝐫(t)=(5,2,3)+t(3,5,5).
Expanding this equation, we get the parametric equations:
x(t)=53t,y(t)=25t,z(t)=3+5t.
Thus, the vector equation for the line is 𝐫(t)=(5,2,3)+t(3,5,5), and the parametric equations for the line are x(t)=53t, y(t)=25t, and z(t)=3+5t.

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