Hatwide k the correct answer is e solution of problem

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hatwide k The correct answer is (E) . Solution of problem 1.3: If −→ F = −→ a + −→ b with −→ a || −→ v and −→ b −→ v , then −→ a is the orthogonal projection of −→ F onto −→ v . From the formula for an orthogonal projection we compute −→ a = proj −→ v −→ F = −→ F · −→ v | −→ v | −→ v = ( 2 , 1 , 3 ) · ( 3 , 1 , 0 ) 3 2 + ( 1) 2 ( 3 , 1 , 0 ) = 5 10 ( 3 , 1 , 0 ) = (bigg 3 2 , 1 2 , 0 )bigg . Therefore −→ b = −→ F −→ a = ( 2 , 1 , 3 ) − (bigg 3 2 , 1 2 , 0 )bigg = (bigg 1 2 , 3 2 , 3 )bigg . 10
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The correct answer is (B) . square Solution of problem 1.4: If we label the points on the plane as P 0 = (1 , 2 , 0), Q 0 = (2 , 2 , 1), and R 0 = (0 , 1 , 1), then we can easily write two vectors parallel to the plane, e.g. vectoru = −−−→ P 0 Q 0 = (2 1) hatwide ı + (2 2) hatwide + (1 0) hatwide k = hatwide ı + hatwide k vectorv = −−−→ P 0 R 0 = (0 1) hatwide ı + (1 2) hatwide + (1 0) hatwide k = hatwide ı hatwide + hatwide k . The normal vector to the plane is given by the cross-product vector n = vectoru × vectorv . We compute vector n = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle hatwide ı hatwide hatwide k 1 0 1 1 1 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = hatwide ı 2 hatwide hatwide k . Thus a point P = ( x, y, z ) on the plane must satisfy the equation vector n · −−→ P 0 P = 0, which is ( x 1) 2( y 2) ( z 0) = 0 or simply x 2 y z = 3 . Substituting the various choices in this equation we see that the only solution is the point (2 , 1 , 3) corresponding to answer (F) . Solution of problem 1.5: The curvature of −→ r ( t ) is given by κ ( t ) = vextendsingle vextendsingle vextendsingle d −→ T dt vextendsingle vextendsingle vextendsingle | −→ r | , where −→ T is the unit tangent vector. We compute −→ r ( t ) = ( 2 , 2 t, 0 ) , and so | −→ r | = 4 + 4 t 2 . In particular we have −→ T = (Big 2(4 + 4 t 2 ) 1 2 , 2 t (4 + 4 t 2 ) 1 2 , 0 )Big . 11
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Substituting in the formula for the curvature we get κ ( t ) = vextendsingle vextendsingle vextendsingle d −→ T dt vextendsingle vextendsingle vextendsingle | −→ r | = vextendsingle vextendsingle vextendsingle d dt (Big 2(4 + 4 t 2 ) 1 2 , 2 t (4 + 4 t 2 ) 1 2 , 0 )Bigvextendsingle vextendsingle vextendsingle (4 + 4 t 2 ) 1 2 = vextendsingle vextendsingle vextendsingle (Big 8 t (4 + 4 2 ) 3 2 , 2(4 + 4 t 2 ) 1 2 8 t 2 (4 + 4 2 ) 3 2 , 0 )Bigvextendsingle vextendsingle vextendsingle (4 + 4 t 2 ) 1 2 Evaluating at t = 0 we get κ (0) = |( 0 , 1 , 0 )| 2 = 1 2 . The correct answer is (D) . square Solution of problem 1.6: The curve of intersection of each surface with the plane x = 2 will be given by the equation in the variables y and z that is obtained from the equation of the surface after the substitution x = 2.
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