Question #159830

Given A= 2i + 4j – 3k , B = 6i-4j+3k , C= 2i-2j+3k

Show that (i) A. B = B.A (ii) A*B = - B*A

(iii) A. (B+C) = A.B + A.C (iv) A* (B+C) = A*B + A*C


1
Expert's answer
2021-01-30T08:17:29-0500

i)


AB=(2,4,3)(6,4,3)=13BA=(6,4,3)(2,4,3)=13=AB\vec{A}\cdot\vec{B}=(2,4,-3)(6,-4,3)=-13\\ \vec{B}\cdot\vec{A}=(6,-4,3)(2,4,-3)=-13=\vec{A}\cdot\vec{B}

ii)


A×B=(2,4,3)×(6,4,3)=(0,24,32)B×A=(6,4,3)×(2,4,3)=(0,24,32)=A×B\vec{A}\times\vec{B}=(2,4,-3)\times(6,-4,3)=(0,-24,-32)\\ \vec{B}\times\vec{A}=(6,-4,3)\times(2,4,-3)=(0,24,32)\\=-\vec{A}\times\vec{B}

iii)


A(B+C)=(2,4,3)(8,6,6)=26AC=(2,4,3)(2,2,3)=13AB=(2,4,3)(6,4,3)=1326=1313\vec{A}\cdot(\vec{B}+\vec{C})=(2,4,-3)(8,-6,6)=-26\\ \vec{A}\cdot\vec{C}=(2,4,-3)(2,-2,3)=-13\\\vec{A}\cdot\vec{B}=(2,4,-3)(6,-4,3)=-13\\-26=-13-13

iv)


A×B=(2,4,3)×(6,4,3)=(0,24,32)A×C=(2,4,3)×(2,2,3)=(6,12,12)A×(B+C)=(2,4,3)×(8,6,6)=(6,36,44)(0,24,32)+(6,12,12)=(6,36,44)\vec{A}\times\vec{B}=(2,4,-3)\times(6,-4,3)=(0,-24,-32)\\\vec{A}\times\vec{C}=(2,4,-3)\times(2,-2,3)=(6,-12,-12)\\\vec{A}\times(\vec{B}+\vec{C})=(2,4,-3)\times(8,-6,6)\\=(6,-36,-44) \\(0,-24,-32)+(6,-12,-12)=(6,-36,-44)




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