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# −→ pr = vextendsingle vextendsingle vextendsingle

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Unformatted text preview: −→ PR = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle hatwide ı hatwide hatwide k − 3 2 − 1 1 − 1 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = (2 − 1) hatwide ı + (3 − 1) hatwide + (3 − 2) hatwide k = hatwide ı + 2 hatwide + 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 , ) 3 2 + ( − 1) 2 ( 3 , − 1 , ) = 5 10 ( 3 , − 1 , ) = (bigg 3 2 , − 1 2 , )bigg . Therefore −→ b = −→ F − −→ a = ( 2 , 1 , − 3 ) − (bigg 3 2 , − 1 2 , )bigg = (bigg 1 2 , 3 2 , − 3 )bigg . 10 The correct answer is (B) . square Solution of problem 1.4: If we label the points on the plane as P = (1 , 2 , 0), Q = (2 , 2 , 1), and R = (0 , 1 , 1), then we can easily write two vectors parallel to the plane, e.g. vectoru = −−−→ P Q = (2 − 1) hatwide ı + (2 − 2) hatwide + (1 − 0) hatwide k = hatwide ı + hatwide k vectorv = −−−→ P R = (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 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 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, ) , 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 , )Big . 11 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 , )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 , )Bigvextendsingle vextendsingle vextendsingle (4 + 4 t 2 ) 1 2 Evaluating at t = 0 we get κ (0) = |( , 1 , )| 2 = 1 2 ....
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−→ PR = vextendsingle vextendsingle vextendsingle...

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