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).and the
x= 3 + 2t
y= 4 - t
z= -1 + t
and finding the angle between them.
We also know that where is the angle between vectors and . 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 and .
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!I don't have a strong linear background.
No, you have dropped a sign: (-1, 6, -3).I believe their intersection of L1 and L2 is at (1,6,-3) (and therefore they intersect).
Not "cross product"- "dot product".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...
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.(Q4)
Let L be the intersection of two planes,
x + y + z = 4
3x - y = 5
(a) find parametric equations for 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>.(b) Find an equation of the plane through the points (1,0,-1) and (2,1,0) that is also parrallel to L.
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.
There is no such thing as "the dot product of the plane and the line"! The dot product is defined for two vectorsFor 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?
Really nothing to do with linear algebra. All vector Calculus.Sorry if these questions are in the wrong area of the forum, they seem like very linear-algebra-related questions in a calculus course.