VAM_44-48

# VAM_44-48 - C Â B = Â 2 2 0 B Â A =(0 Â 2 0 = n =(0 4 The...

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Tutorial VAM: Vector Algebra and an Introduction to Matrices Row 3 by 2 = 39 : 0 @ 4 0 0 0 ¡ 1 2 0 0 0 39 2 1 A ! 0 @ 1 0 0 0 1 0 0 0 1 1 A = ) 0 @ 20 13 ¡ 4 39 ¡ 40 39 2 13 ¡ 1 13 ¡ 7 26 ¡ 3 2 4 1 1 A ! 0 @ 5 13 ¡ 1 39 ¡ 10 39 ¡ 4 13 2 13 7 13 ¡ 1 13 8 39 2 39 1 A which = B ¡ 1 ; as can be checked by multiplying. Exercise 43. We have j B j = ¡ 39 : Computing the minors, we have ¯ ¯ B (11) ¯ ¯ j B j = ¡ 15 ¡ 39 = 5 13 ; ¡ ¯ ¯ B (21) ¯ ¯ j B j = ¡ ¡ 4 +3 ¡ 39 = ¡ 1 39 ; ¯ ¯ B (31) ¯ ¯ j B j = 10 ¡ 39 = ¡ 10 39 ¡ ¯ ¯ B (12) ¯ ¯ j B j = ¡ ¡ 2 ¡ 10 ¡ 39 = ¡ 4 13 ; ¯ ¯ B (22) ¯ ¯ j B j = ¡ 8+2 ¡ 39 = 2 13 ; ¡ ¯ ¯ B (32) ¯ ¯ j B j = ¡ 20 +1 ¡ 39 = 7 13 ¯ ¯ B (13) ¯ ¯ j B j = 3 ¡ 39 = ¡ 1 13 ; ¡ ¯ ¯ B (23) ¯ ¯ j B j = ¡ 12 ¡ 4 ¡ 39 = 8 39 ; ¯ ¯ B (33) ¯ ¯ j B j = ¡ 2 ¡ 39 = 2 39 Exercise 44. n = ( C ¡ B ) £ ( B ¡ A ) = C £ B ¡ C £ A ¡ B £ B + B £ A = C £ B + A £ C + B £ A = ) n ¢ ( r ¡ A )= r ¢ ( C £ B + A £ C + B £ A ) ¡ A ¢ ( C £ B ) : Applying Eqn. 63 to each of the triple products results in Eqn. 71. Exercise 45. The minimum distance from the origin to the plane will be in the direction of n , since that line intersects the plane perpendicularly. The projection along that line of any vector A directed from the origin to the plane willbe n ¢ A = j n j = ^n ¢ A : Exercise 46. We can use Eqn. 71 directly, or we can compute n and use Eqn. 70. For these vectors, the latter approach is the more efficient.
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Unformatted text preview: C Â¡ B =( Â¡ 2 ; 2 ; 0); B Â¡ A =(0 ; Â¡ 2 ; 0) = ) n =(0 ; ; 4) : The equation for the plane is thus 4( z Â¡ 1) = 0 or z = 1 : The distance of this plane to the origin is ^n Â¢ A = ^z Â¢ A = A z = 1 . Exercise 47. The minimum distance from the origin to the line will be in the plane of A and B and perpendicular to the line. This defines the direction of n 2 : The projection of A or B along this direction is ^n 2 Â¢ A = ^ n 2 Â¢ B : Exercise 48. n 1 = Â¯ Â¯ Â¯ Â¯ Â¯ Â¯ ^x ^y ^ z 2 Â¡ 1 3 1 2 1 Â¯ Â¯ Â¯ Â¯ Â¯ Â¯ = Â¡ 7 ^x + ^y +5 ^ z : n 2 = Â¯ Â¯ Â¯ Â¯ Â¯ Â¯ ^ x ^y ^ z Â¡ 7 1 5 1 Â¡ 3 2 Â¯ Â¯ Â¯ Â¯ Â¯ Â¯ = 17 ^ x +19 ^y +20 ^z : The constraint equations, (74), give n 1 Â¢ r = Â¡ 7 x + y + 5 z = 0; 49...
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