By Daniel Zelinsky and Samuel S. Saslaw (Auth.)
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Extra info for A First Course in Linear Algebra
D) V · ( ( w + w ' ) + w " ) = V · ( w + w ' ) V · w + ν · w ' + V · w " , and similarly for ( e ) . 6 + ν · = THEOREM 6, c ] [a, (ai + 6j PROOF. (ai + ck) . [α', . (a'i + ck) . (a'i + = V . = αα' + hV + cc' c'k) = αα' + hV + cc'. 4(d) with + h]+ 6', c ' ] ν = a i - h 2>j + 6'j + (a'i) ck we get c'k) + V . (6'j) + V . (c'k). 4(e) to get ν · ( a ' i ) = a i · a ' i + 6j · a ' i + c k · a ' i and do the same with the other two dot products, finally getting the whole dot product expressed as a smn of nine terms (ai) .
PM = ^PQ and PQ is equivalent to w ~ ν so PM and " P M " are equivalent to i(w — v ) . But the last two of these three vectors have their tails at 0 , so they are equal vectors. Therefore, OM = OP + "PM" = V + | ( w - v ) = V + - i v = | v+ i w = M v + w ) . 4 follows immediately. Let the end points of the line segment in question be Ρ and Q and let ν = [α, bj c ] and w = [α', ö', c'] be the position vectors of Ρ and Qj respectively. 4, with 0 the origin. The coordinates of the midpoint Μ will be the components of OM = i ( v + w ) = hila, 6, c ] + [α', 6', c']) = [ ( α + α')/2, (6 + 6 ' ) / 2 , (c + c ' ) / 2 ] .
Solve these for a, 6, c, d. ). The best we can hope for is to solve for three unknowns 58 2. Planes and Lines in terms of the fom-th. B y subtracting, we eliminate d from the first and second equations, then from the first and third. The original equations are then equivalent to (that is, have the same solutions as) a + c - d = 0 α + 6 + 2c =0 5a ~ 6 + 2c =0. Then eliminate c from the last two, and get another equivalent system α a+ + c - d = 0 b + 2c 4a - 2& =0 = 0, which says 6 = 2a, c = - (a + 6 ) / 2 = ( - 3 / 2 ) a , d = α + c = —|a.