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MSU MATH 234 HWi Answer Section 12.1 (52). Find the center and radius of the sphere 3a? + By? + 32? 4 2y—22=9 Solution: 2 2 2 2 2 2 2 2 Ba? + By? 4322 + 2y—22=9 me a tyr ta? oy Fe=3 1 1 2 2 = 2 92 = +(v +23u) + (2 232) 3 1 1\? i 1\? ry? f1\? 2 2 =, a 2 9s _ =3 a = x +(v +250+ (3) )+(- 2++(3)) 3+(3) +(5) 2 2 ay? 1\? _ 83 _ ( v88 (z of+(v+3) +(¢ 3) =o => The center is (0, =, }) and the radius is V3. Section 12.2 [8]. Let w= (8,—-2) and v = (-2,5). Find the (a) component form and (b) magnitude of the vector. Solution: (3) ut pe ay G2) + C29) = (GE) + EB) = GEE a5 39 By (2 v (©) |agut Hel = [(52, B)| = V GR) +B)" = [34]. Find a vector of magnitude 3 in the direction opposite to the direction of v = 1i— hi - Be Solution: lel = (2) + 4)? +G y= The direction of v is pu = Jy The opposite direction is — rr The desired vector is 3 (- a’) = 3(4i + Ji + 3) = —V3i + V/8j + V8K ty at ad — yeh Section 12.3 [3]. Let v= 1084+ 11j-2k and u=3j+4k Find (a) v-, Jul, [ul (b) the cosine of the angle between v and « (c) the scalar component of wu in the direction of v (d) the vector projyu Solution: (a) v-u=10-04+11-3-2-4=25 ol = yf 10? + 112 + (—2)? = 15 |e] = /0? +3? + 4? = 5 = tee I (b) cos = tt = BE =D 255 (2 tr == 3 (2) projyu = feu = yy (108 + 11j ~ 2k) = Pi+ Yj 2h 
