Solutions to exercises on vector calculus, including calculations of volumes, parametric representations, and distances between points and planes.
Extract
[...] Exam n°4 Exercise On seeks such as . On wants to solve the system here is equivalent to the last equation is never verified so the entire set of solutions is . Exercise By definition of the vector product, we have the following relation . Or, it is known that according to the statement . It is deduced that . It is also known by definition of the vector product that . The first expression gives and the second tells us that finally . [...]
[...] Exercise On a thus . On a , thus (also For it's immediate and . On a so: Then: Exercise One notes the components of the matrix . One notes successively the implications to the different affirmations: On en déduit directement que . Exercise On a the characteristic polynomial associated to here is worth . The three eigenvalues are therefore directly . The matrix is a matrix whose definition I do not know. 3. Exam Exercise " " On a so for equality we need . [...]
[...] The parametric equations are: . En isolant and on a et , then by injection we get the Cartesian equation of the beam which is then so it is still . If a point belongs to the line then its coordinates verify so it is injected into the Cartesian equation of the plane, or even so . The intersection is a point of coordinate . The line a for the direction vector the vector , in addition to the point belongs to this line so the distance between a point and the slope is. [...]
[...] If the point belongs to this plane, then it must check the Cartesian equation which gives so . The Cartesian equation of the sought plane is therefore . Exercise On a so . Exercise Exactly the same question as the of the exercise 4 of the homework 1. On One then obtains the matrix next: Like the column vectors of are the vectors of an orthonormal basis, we know that the matrix is orthogonal and that (same question as the of exercise 4 of homework 1). [...]
[...] Exam n°2: Exercise By definition . In addition On deduce from the second equality that and then with the first that . Finally, . On saw that for this it is necessary so it and . On can therefore take any. We therefore have a parametric system of equations . On a donc the Cartesian equation system of the line which is . By construction, this area is equal to the norm of the vector product of the two vectors, which in this case . [...]
