Consider the following two lines: L1 : x =(y - 1)/2=(z - 1)/3 L2 : 6(x - 1) = 3y + 1 = z. (a) Show that L1 and L2 are skew. Document Preview: Name: Solutions to Practice Midterm1.(a) Are the...

Name: Solutions to Practice Midterm 1. (a) Are the vectorsh1; 5; 3i andh3; 0; 2i perpendicular? (b) Are the points P = (1;1; 1), Q = (2; 1;1), R = (0; 3;1), S = (1; 0; 1) coplanar? (c) Find the area of the triangle with vertices A = (1; 0; 0), B = (0; 1; 0), C = (0; 0; 2). Solution. (a) Their dot product is 3 + 0 + 6 = 3; which is not zero, therefore they are not perpendicular. ! ! ! ~ (b) We just need to check if the triple productj~a
(b~c)j of the vectorsPQ,PR andPS is zero. ! ! ! ! ! We havePQ =h1; 2;2i,PR =h1; 4;2i,PS =h2; 1; 0i, and soPRPS =h2; 4; 7i and ! ! ! PQ
(PRPS) =4: Since this is not zero, the 4 points are not coplanar. ! (c) The area of the triangle is half the area of the parallelogram spanned by the vectors AB = ! h1; 1; 0i and AC =h1; 0; 2i. This area is the length of the vector ! ! ABAC =h2; 2; 1i; p which is 4 + 4 + 1 = 3, so the area of the triangle is 3=2. 2. (a) Find an equation for the plane through the points P = (1; 0; 1), Q = (1; 2; 2), R = (3; 1; 2). (b) Find the intersection of the line ~r(t) =h1; 1; 0i +th3; 1; 1i with the plane x +y +z = 1. Solution. (a) We need to nd 2 things: a vector ~n normal to the plane, and a point on the plane. To do ! ! ! ! this, we can just take~n =PQPR. We have PQ =h0; 2; 1i, PR =h2; 1; 1i, so ! ! ~n =PQPR =h1; 2;4i: Since the plane passes through P = (1; 0; 1), we take this to be our point, and therefore the plane has equation (x 1) + 2y 4(z 1) = 0: 1(b) Write the line as x = 1 + 3t; y = 1 +t; z =t: To nd the intersection points we substitute these into the equation for the planex+y+z = 1 and solve for t: 1 + 3t + 1 +t +t = 1; so t =1=5, and the intersection point is (1 3=5; 1 1=5;1=5) = (2=5; 4=5;1=5): 3. Consider the following two lines: y 1 z 1 L : x = = 1 2 3 L : 6(x 1) = 3y + 1 =z: 2 (a) Show that L and L are skew. 1 2 Solution. (a) We can get a parametric equation for the line L by setting all the components equal to t: 1 t =x = (y 1)=2...