Question

Vector (Cross) Product 1. Find the vector product (2j-2k) x 5k. Sketch all three vectors onto the coordinate system below Answer: 10 Find the vector product of i+4j-3k and -2i+j-5k. Prove that your answer is perpendicular to the first two vectors by using the dot product Answer: -17i+11j+9k or 17i-11j-9k, depending on the order in which you took the cross product. 2.

Answer 1

1)

Vector product of given vectors is

$(2\bold{j}-2\bold{k})\times5\bold{k}=10(\bold{j}\times\bold{k})-10(\bold{k}\times\bold{k})=10(\bold{i})-10(0)=10\bold{i}$

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2)

Vector product of given vectors is

${(\bold{i}+4\bold{j}-3\bold{k})\times(-2\bold{i}+\bold{j}-5\bold{k})=-2(\bold{i}\times\bold{i})+(\bold{i}\times\bold{j})-5(\bold{i}\times\bold{k})-8(\bold{j}\times\bold{i})+4(\bold{j}\times\bold{j})-20(\bold{j}\times\bold{k})+6(\bold{k}\times\bold{i})-3(\bold{k}\times\bold{j})+15(\bold{k}\times\bold{k})}$

${(\bold{i}+4\bold{j}-3\bold{k})\times(-2\bold{i}+\bold{j}-5\bold{k})=-2(0)+(\bold{k})-5(-\bold{j})-8(-\bold{k})+4(0)-20(\bold{i})+6(\bold{j})-3(-\bold{i})+15(0)}$

${(\bold{i}+4\bold{j}-3\bold{k})\times(-2\bold{i}+\bold{j}-5\bold{k})=(\bold{k})+5(\bold{j})+8(\bold{k})-20(\bold{i})+6(\bold{j})+3(\bold{i})$

$(\bold{i}+4\bold{j}-3\bold{k})\times(-2\bold{i}+\bold{j}-5\bold{k})=-17\bold{i}+11\bold{j}+9\bold{k}$

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The resultant vector is $-17\bold{i}+11\bold{j}+9\bold{k}$

Dot products of the resultant vector with the first vector is ${(\bold{i}+4\bold{j}-3\bold{k})\cdot(-17\bold{i}+11\bold{j}+9\bold{k})=-17(\bold{i}\cdot\bold{i})+11(\bold{i}\cdot\bold{j})+9(\bold{i}\cdot\bold{k})-68(\bold{j}\cdot\bold{i})+44(\bold{j}\cdot\bold{j})+36(\bold{j}\cdot\bold{k})+51(\bold{k}\cdot\bold{i})+33(\bold{k}\cdot\bold{j})-27(\bold{k}\cdot\bold{k})}$ ${(\bold{i}+4\bold{j}-3\bold{k})\cdot(-17\bold{i}+11\bold{j}+9\bold{k})=-17(1)+11(0)+9(0)-68(0)+44(1)+36(0)+51(0)+33(0)-27(1)}$

${(\bold{i}+4\bold{j}-3\bold{k})\cdot(-17\bold{i}+11\bold{j}+9\bold{k})=-17+44-27=0$

Dot product of the vectors is zero, hence they are perpendicular to each other.

Dot products of the resultant vector with the second vector is ${(-2\bold{i}+\bold{j}-5\bold{k})\cdot(-17\bold{i}+11\bold{j}+9\bold{k})=34(\bold{i}\cdot\bold{i})-22(\bold{i}\cdot\bold{j})-18(\bold{i}\cdot\bold{k})-17(\bold{j}\cdot\bold{i})+11(\bold{j}\cdot\bold{j})+9(\bold{j}\cdot\bold{k})+85(\bold{k}\cdot\bold{i})-55(\bold{k}\cdot\bold{j})-45(\bold{k}\cdot\bold{k})}$

${(-2\bold{i}+\bold{j}-5\bold{k})\cdot(-17\bold{i}+11\bold{j}+9\bold{k})=34(1)-22(0)-18(0)-17(0)+11(1)+9(0)+85(0)-55(0)-45(1)}$$(-2\bold{i}+\bold{j}-5\bold{k})\cdot(-17\bold{i}+11\bold{j}+9\bold{k})=34+11-45=0$

Dot product of the vectors is zero, hence they are perpendicular to each other.

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