The subject is linear algebra 3.5.A Find the intersection of the two lines (x, y

Question

The subject is linear algebra

3.5.A Find the intersection of the two lines (x, y, z)- (5,-3,4) + t(3,-2,1) and ( x, y, z)- (3,-4,5) + s(-1,3,-2). Note that s and t might be different numbers at the point of intersection. If you make any arithmetic mistake you will cause the two lines to not intersect. 3.5.B Find the equation of the plane that contains the two intersecting lines ( x, y, z) = (5,3,4) + t(3,-2,1) and ( x, y, z) = (3,4,5) + s(-1,3,2). Find both the vector form of the equations and the point normal form. The previous problem is very helpful.

Explanation / Answer

3.5 A.

The 1st and the 2nd lines intersect if the equations 5+3t = 3-s or, s+3t= -2…(1) , -3-2t = -4+3s or, 3s+2t = 1…(2) and 4+t = 5-2s or, 2s+t = 1… (3) are consistent.   The augmented matrix of this linear system is A =

1

3

-2

3

2

1

2

1

1

The RREF of A is

1

0

1

0

1

-1

0

0

0

It implies that there is a unique solution i.e. s =1 and t= -1, to the above linear system, so that the given 2 lines intersect.The point of intersection is (x,y,z) = (3,-4,-5)+1(-1,3,-2) = (2-1,-7).

B. The direction vector of the 2 lines are v1 = (3,-2,1) and v2 = (-1,3,-2) respectively. Hence the normal to the plane is v1 x v2 = (1,5,11). Further, the point of intersection of the 2 lines, i.e. (2,-1,-7) lies on the plane. Hence the point normal form of the equation of the plane is (1,5,11).(x-2,y+1,z+7) = 0. It can easily be converted into cartesian form.

1

3

-2

3

2

1

2

1

1

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