How do you find the parametric equations of the line that passes through <mstyle displaystyle="t

Jase Howe

Jase Howe

Answered question

2022-06-21

How do you find the parametric equations of the line that passes through A = ( 7 , - 2 , 4 ) and that is parallel to the line of intersection of the planes 4 x - 3 y - z - 1 = 0 and 2 x + 4 y + z - 5 = 0 ?

Answer & Explanation

Cristopher Barrera

Cristopher Barrera

Beginner2022-06-22Added 24 answers

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The plane can be represented by
Π → 〈 p - p 0 , n → 〉 = 0 or
( x - x 0 ) a + ( y - y 0 ) b + ( z - z 0 ) c = 0 or
a x + b y + c z - a x 0 - b y 0 - c z 0 = 0
Here p = ( x , y , z ) , p 0 = ( x 0 , y 0 , y 0 ) and n → = ( a , b , c )
So
Π 1 → 〈 p - p 1 , n → 1 〉 = 0 and
Π 2 → 〈 p - p 2 , n → 2 〉 = 0
with
p 1 = ( 0 , 0 , - 1 ) and n → 1 = ( 4 , - 3 , - 1 )
p 2 = ( 0 , 0 , 5 ) and n → 2 = ( 2 , 4 , 1 )
Let
L i → p = p i + λ v → the intersection line between the two planes
If L i = Π 1 ∩ Π 2 then
〈 p i - p 1 + λ v → , n → 1 〉 = 0 and
〈 p i - p 2 + λ v → , n → 2 〉 = 0
or
〈 p i - p 1 , n → 1 〉 + λ 〈 v → , n → 1 〉 = 0
〈 p i - p 1 , n → 2 〉 + λ 〈 v → , n → 2 〉 = 0
but p i ∈ Π 1 ∩ Π 2 so
〈 p i - p 1 , n → 1 〉 = 0 and
〈 p i - p 1 , n → 2 〉 = 0
then
v → = μ n → 1 × n → 2 because it must be orthogonal to both.
So, the line parallel to L i passing by A is
L A → A + γ n → 1 × n → 2
n 1 × n 2 = ( 1 , - 6 , 22 ) so the parametric equations are
{ x = 7 + γ y = - 2 - 6 γ z = 4 + 22 γ
Kyla Ayers

Kyla Ayers

Beginner2022-06-23Added 8 answers

Step 1
Look at my Solution, based upon more of Algebra , and a little bit of Geometry!
Let the given planes be
π 1 : 4 x - 3 y - z - 1 = 0 ... ( i ) , & , π 2 : 2 x + 4 y + z - 5 = 0 ... ( i i )
( i ) + ( i i ) ⇒ 6 x + y - 6 = 0 , i . e . , y = 6 - 6 x ... ... ... ( 1 )
Sub.ing in (i).
4 x - 3 ( 6 - 6 x ) - z - 1 = 0 , i . e . , 22 x - 19 = z ... ... ... ( 2 )
Altogether, we have,
y = 6 - 6 x , z = 22 x - 19 , 7 , of course, x = x
Since the above have been derived by solving π 1 & π 2 , this means that,
∀ x ∈ ℝ , ( x , 6 - 6 x , 22 x - 19 ) ∈ π 1 ∩ π 2 , or ,
π 1 ∩ π 2 = { ( 0 , 6 , - 19 ) + x ( 1 , - 6 , 22 ) : x ∈ ℝ }
Observe that, since the reqd. line is ∣ ∣ to π 1 ∩ π 2 the direction of the reqd. line is (along) the vector ( 1 , - 6 , 22 )
Finally, using the pt. A ( 7 , - 2 , 4 ) on the reqd. line, we get, its eqn.
r → = ( x , y , z ) = ( 7 , - 2 , 4 ) + t ( 1 , - 6 , 22 ) , t ∈ ℝ

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