Equation of a plane containing a line and parallel to a vector

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To find the equation of a plane containing the line r = i + 2j + 3k + t(i - j + k) and parallel to the vector v = -i + 2j + 3k, the correct approach involves identifying a direction vector from the line and taking the cross product with vector v. The direction vector of the line can be derived from its parametric equation, while the cross product will yield a normal vector to the plane. This normal vector, along with a point on the line, can then be used to formulate the equation of the plane. The discussion emphasizes the importance of correctly identifying vectors and their relationships to derive the plane's equation.
skyflashings
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Hi! Any help would be appreciated!

Give the equation of a plane containing the line r = i + 2j + 3k + t(i - j + k) and is parallel to vector v = -i + 2j + 3k.


I'm using this as reference.


My guess is to treat the problem similar to finding an equation of a plane which contains a line and passes through a point. So, I would get a vector on the line (say from t = 0 and t = 1), and a vector from a point on the line (say t = 0) to the vector v (since it is parallel, any multiple of (-1, 2, 3) will suffice). Using these two vectors get the normal, which we can use to derive the equation of the plane.

Is this approach to the solution correct or is there more to it than that? I believe I may be oversimplifying the problem in regards to the parallel vector v.
 
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skyflashings said:
Hi! Any help would be appreciated!

Give the equation of a plane containing the line r = i + 2j + 3k + t(i - j + k) and is parallel to vector v = -i + 2j + 3k.


I'm using this as reference.


My guess is to treat the problem similar to finding an equation of a plane which contains a line and passes through a point. So, I would get a vector on the line (say from t = 0 and t = 1), and a vector from a point on the line (say t = 0) to the vector v (since it is parallel, any multiple of (-1, 2, 3) will suffice).
I don't think the part about getting a vector from a point on the given line to the vector v will work. Plus, I'm not even sure what you mean by getting a vector from a point to another vector.

The most straightforward way to approach this problem, I believe, is to find a vector that has the same direction as the given line, and then take the cross product of that vector and v. That will give you a third vector that is perpendicular to both vectors, that will serve as a normal to the plane. Once you have that vector, all you need is a point on the plane, and you can write the equation of the plane.
skyflashings said:
Using these two vectors get the normal, which we can use to derive the equation of the plane.

Is this approach to the solution correct or is there more to it than that? I believe I may be oversimplifying the problem in regards to the parallel vector v.
 
Ah, yes you are right. I wasn't thinking of taking the cross between the vector and a vector 'in the direction of' the line. Thanks very much!
 
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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