A plane pi contains the line r = i + 3j - k +lambda (2i - j + k). If pi is parallel to 3i - 4j, how to find the vector equation of pi?

Kareem Mejia

Kareem Mejia

Answered question

2022-11-05

A plane π contains the line r = i + 3 j k + λ ( 2 i j + k ). If π is parallel to 3i - 4j, how to find the vector equation of π?
I know that parallel planes have the same perpendicular vector. But I don’t seem to see how to find any normal vectors to the plane. I can’t cross product two lines. How should I solve this question?

Answer & Explanation

klofnu7c2

klofnu7c2

Beginner2022-11-06Added 14 answers

Both the vectors 3i−4j and 2i−j+k are parallel to the plane, so the normal vector of the plane is perpendicular to both of these vectors. We can find it using the cross product
r = i j k 3 4 0 2 1 1 = 4 i 3 j + 5 k
This gives us the equation of the plane as −4x−3y+5z=c or 4x+3y−5z=d. The point i+3j−k lies in this plane, so substituting these values gives us the value of d.
d = 4 9 5 = 18
The equation of the plane is thus 4 x 3 y + 5 z = 18

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