Given that A = 2i + 3j - k, B = i - j +2k and C = 3i + 4j + k find:
a) A+2B
b) |A- B+2C|
c) D such that A - B + C 3D = 0
A=(2,3,−1),B=(1,−1,2),C=(3,4,1)a:A+2B=(2+2,3−2,−1+4)=(4,1,3)b:A−B+2C=(2−1+6,3+1+8,−1−2+2)=(7,12,−1)∣A−B+2C∣=72+122+12=194c:A−B+C+3D=0⇒⇒D=13(B−A−C)=13(1−2−3,−1−3−4,2+1−1)==(−43,−83,23)A=\left( 2,3,-1 \right) ,B=\left( 1,-1,2 \right) ,C=\left( 3,4,1 \right) \\a:\\A+2B=\left( 2+2,3-2,-1+4 \right) =\left( 4,1,3 \right) \\b:\\A-B+2C=\left( 2-1+6,3+1+8,-1-2+2 \right) =\left( 7,12,-1 \right) \\\left| A-B+2C \right|=\sqrt{7^2+12^2+1^2}=\sqrt{194}\\c:\\A-B+C+3D=0\Rightarrow \\\Rightarrow D=\frac{1}{3}\left( B-A-C \right) =\frac{1}{3}\left( 1-2-3,-1-3-4,2+1-1 \right) =\\=\left( -\frac{4}{3},-\frac{8}{3},\frac{2}{3} \right)A=(2,3,−1),B=(1,−1,2),C=(3,4,1)a:A+2B=(2+2,3−2,−1+4)=(4,1,3)b:A−B+2C=(2−1+6,3+1+8,−1−2+2)=(7,12,−1)∣A−B+2C∣=72+122+12=194c:A−B+C+3D=0⇒⇒D=31(B−A−C)=31(1−2−3,−1−3−4,2+1−1)==(−34,−38,32)
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