MATH 126 Midterm: MATH 126 UW Midterm 1 Fall 10duchampExIans

• School: UW
• Department: Mathematics
• Course: MATH 126
• Professor: All

Document Summary

Solution. (a) note that ab = i j and ac = i 3 j 4 k. Let be the angle at vertex a. Ab|| (b) let n = ab (c) since n is normal to the plane and a is on the plane, setting r = x i + y j + z k and r0 = i + 2 j + 3 k, in the. 2 equation n (r r0) = 0 yields the equation 4x 4y + 4z 8 = 0 or x y + z 2 (d) distance = (cid:12)(cid:12)(cid:12)(cid:12) Hence sin( ) = cos( ). for t. this gives t = 1. Since r(1) = i + j k, q = (1, 1, 1) . = sin 1(1/ 6) . (note: = /2 cos 1(1/ 6) is another way to write this. )