# Linear Algerba in Calculus

• Jan 20th 2010, 05:32 PM
matt.qmar
Linear Algerba in Calculus
Hello!

We got an assignment in a Calculus class that appears to me to be very linear algebra-ish... I can do some of it by intution but I have two questions,

(Q1)
To show the

line (L1):
containing the points (3,10,5) and (8,15,15)

and the

line (L2):
x= 3 + 2t
y= 4 - t
z= -1 + t

intersect,
and finding the angle between them.

I don't have a strong linear background. I believe their intersection of L1 and L2 is at (1,6,-3) (and therefore they intersect).
I am not sure how to begin on finding the angle between them. Does the cross product have an application here? Not sure exactly what the cross product represents, so I am not sure how to use it, but I know it involved the sine of the angle between the two lines...

(Q4)
Let L be the intersection of two planes,

x + y + z = 4

and

3x - y = 5

(a) find parametric equations for L
(b) Find an equation of the plane through the points (1,0,-1) and (2,1,0) that is also parrallel to L.

For part (a), I am assuming L represents a line, however I am not sure how to get L in standard form, let alone parametrically.
also in part (b) I think I know that the dot product of the plane and the line will be 0, if the plane is parrallel to the line?

Sorry if these questions are in the wrong area of the forum, they seem like very linear-algebra-related questions in a calculus course.

Thanks!
• Jan 21st 2010, 07:18 AM
HallsofIvy
Quote:

Originally Posted by matt.qmar
Hello!

We got an assignment in a Calculus class that appears to me to be very linear algebra-ish... I can do some of it by intution but I have two questions,

(Q1)
To show the

line (L1):
containing the points (3,10,5) and (8,15,15)

The vector from (3, 10, 5) to (8, 15, 15) is <8- 3, 15-10, 15- 5>= <5, 5, 10> so the line is given by x= 3+ 5s, y= 10+ 5s, z= 5+ 10s. Notice that when t= 0 that gives (3, 10, 5) and when t= 1 it gives (8, 15, 15).

Quote:

and the

line (L2):
x= 3 + 2t
y= 4 - t
z= -1 + t

intersect,
and finding the angle between them.
This line has "direction vector" <2, -1, 1> (from the coefficients of t). In order that they intersect at (x,y,z) we must have x= 3+ 5s= 3+ 2t, y= 10+ 5s= 4- t, and z= 5+ 10s= -1+ t. The first equation says 5s= 2t or s= (2/5)t. Putting that into 10+ 5s= 4- t gives 10+ 2t= 4- t so 3t= -6 and t= -2. Then s= (2/5)t= -4/5. x= 3+ 5s= 3+5(-4/5)= -1 and y= 10+ 5s= 10+5(-4/5)= 6. But we must also check that the z equations are satisfied. Putting those into the z equations, z= 5+ 10s= 5+ 10(-4/5)= 5- 8= -3 and z= -1+ t= -1-2= -3. Since those are the same, the two lines intersect at (-1, 6, -3).

We also know that $\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)$ where $\theta$ is the angle between vectors $\vec{u}$ and $\vec{v}$. Here, the dot product of the two vectors, <5, 5, 10> and <2, -1, 1> is 5(2)- 5+ 10= 10- 5+ 10= 15 and their lengths are $\sqrt{150}= 5\sqrt{6}$ and $\sqrt{6}$.

Quote:

I don't have a strong linear background.
Then that's why you think these are "linear algebra"! Just because they deal with vectors doesn't mean they have very much to do with linear algebra!

Quote:

I believe their intersection of L1 and L2 is at (1,6,-3) (and therefore they intersect).
No, you have dropped a sign: (-1, 6, -3).

Quote:

I am not sure how to begin on finding the angle between them. Does the cross product have an application here? Not sure exactly what the cross product represents, so I am not sure how to use it, but I know it involved the sine of the angle between the two lines...
Not "cross product"- "dot product".

Quote:

(Q4)
Let L be the intersection of two planes,

x + y + z = 4

and

3x - y = 5

(a) find parametric equations for L
From the second equation, y= 3x+ 5. Putting that into the first equation gives x+ 3x+ 5+ z= 4 or z= -4x- 1. Use x itself as the parameter (but I'm going to "rename" it t): x= t, y= 3t+ 5, z= -4t-1.

Quote:

(b) Find an equation of the plane through the points (1,0,-1) and (2,1,0) that is also parrallel to L.
Any plane containing (x0, y0, z0) can be written as A(x- x0)+ B(y- y0)+ C(z- z0)= 0 for some numbers, A, B, C. Further, the vector perpendicular to the plane is <A, B, C>. In order that the plane be parallel to L, we must have <A, B, C> perpendicular to <1, 3, -4>.
Because (1, 0, -1) is in the plane, we can write its equation as A(x-1)+ B(y- 0)+ C(z+1)= 0. Because it is perpendicular to <1, 3, -4> we must have the <A, B, C>.<1, 3, -4>= A+ 3B- 4C= 0. Because (2, 1, 0) is in the plane, we must have A(2-1)+ B(1-0)+ C(0+1)= A+ B+ C= 0. That gives two equations for A, B, C but because we could multiply the entire equation by a constant without changing the plane so we could take any one of them to be, say, 1, and solve for the other two.

Quote:

For part (a), I am assuming L represents a line, however I am not sure how to get L in standard form, let alone parametrically.
also in part (b) I think I know that the dot product of the plane and the line will be 0, if the plane is parrallel to the line?
There is no such thing as "the dot product of the plane and the line"! The dot product is defined for two vectors

Quote:

Sorry if these questions are in the wrong area of the forum, they seem like very linear-algebra-related questions in a calculus course.

Thanks!
Really nothing to do with linear algebra. All vector Calculus.