Equation of plane given two parallel lines?

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To find the equation of a plane containing two parallel lines, the first step involves converting the given lines into parametric form. The cross product of the direction vectors from the lines yields a normal vector, which is essential for the plane's equation. A point on the plane can be identified from the intersection of the lines, and another point can be derived from the parametric equations. The discussion highlights a correction in the problem statement regarding the parameterization of one line, which affects the calculations for the vector PQ used to determine the plane's equation. Ultimately, the correct normal vector and points lead to the desired plane equation.
Isaac Wiebe
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Homework Statement


Find the equation that contains the lines:
2x + 3y + 4z = 6, x - 2y + z = 3

and

(2x - 1)/22 = (y + 2)/2 = (1 - z)/3


Homework Equations



A plane (in point normal form) is defined by a point on the plane and a vector normal to it.
In general, the form of A(x - x0) + B(y - y0) + C(z - z0) = 0.

You can use the cross product to determine a normal vector given multiple lines on the plane.


The Attempt at a Solution



Step 1: Convert L1 to parametric form. Take the cross product of <2, 3, 4> and <1, -2, 1>
and that is the vector <11, 2, -7>. A point that is on both planes forming the line of intersection is (3, 0, 0). Thus L1: x = 3 + 11T, y = 2T, and z = -7T.

Step 2: Convert L2 to parametric form. x = 11T + 0.5, y = 2T -2 and z = 1 - 7T.

Since these lines are indeed parallel, we have to find another line or vector on the plane. Consider vector PQ, where point P (3, 0, 0) is on L1 and point Q (0.5, -2, 1) is on L2. Consequently, PQ is <-3.5, -2, 1>. The normal vector is found by taking the cross product of PQ and L1, and I am having a difficult time obtaining the correct answer, 24x - 13y + 34z = 72.
 
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Isaac Wiebe said:

Homework Statement


Find the equation that contains the lines:
2x + 3y + 4z = 6, x - 2y + z = 3

and

(2x - 1)/22 = (y + 2)/2 = (1 - z)/3


Homework Equations



A plane (in point normal form) is defined by a point on the plane and a vector normal to it.
In general, the form of A(x - x0) + B(y - y0) + C(z - z0) = 0.

You can use the cross product to determine a normal vector given multiple lines on the plane.


The Attempt at a Solution



Step 1: Convert L1 to parametric form. Take the cross product of <2, 3, 4> and <1, -2, 1>
and that is the vector <11, 2, -7>. A point that is on both planes forming the line of intersection is (3, 0, 0). Thus L1: x = 3 + 11T, y = 2T, and z = -7T.

Step 2: Convert L2 to parametric form. x = 11T + 0.5, y = 2T -2 and z = 1 - 7T.

I don't get that z.
 
Oops, the original problem read (1 - z) / 7, NOT (1 - z) / 3. My mistake!
 
Now check your calculations for PQ.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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