Question #221572

1)If P>Q =4i+3j+2k
Find the magnitude of PQ' i.e /PQ/
(2) if A'=3i-j+2k and'B=2i+3j-k, Find
(1) 'A×B'
(2) A'×(2A'+3B')
(3)A'.B'

Expert's answer

Assume

P=4i^+3j^+2k^P=4\hat{i}+3\hat{j}+2\hat{k}

Q=2i^+3j^+k^Q=2\hat{i}+3\hat{j}+\hat{k}

∣P∣=16+9+4=30∣Q∣=4+9+1=14|P|=\sqrt{16+9+4}=\sqrt{30}\\|Q|=\sqrt{4+9+1}=\sqrt{14}

P>QP>Q correct statement

PQ=(4i^+3j^+2k^).(2i^+3j^+k^)(4\hat{i}+3\hat{j}+2\hat{k}).(2\hat{i}+3\hat{j}+\hat{k})

PQ=8+9+2=19

2.

A=3i^−j^+2k^B=2i^+3j^−k^A=3\hat{i}-\hat{j}+2\hat{k}\\B=2\hat{i}+3\hat{j}-\hat{k}

A×B=∣i^j^k^3−1223−1∣A\times B=\begin{vmatrix} \hat{i} & \hat{j}&\hat{k} \\ 3& -1&2\\2&3&-1 \end{vmatrix}


A×B=i^(1−6)−j^(−3−4)+k^(9+2)A\times B=\hat{i}(1-6)-\hat{j}(-3-4)+\hat{k}(9+2)

A×B=−5i^+7j^+11k^A\times B=-5\hat{i}+7\hat{j}+11\hat{k}

(2)


2A+2B=2×(3i^−j^+2k^)+2×(2i^+3j^−k^)2A+2B=2\times(3\hat{i}-\hat{j}+2\hat{k})+2\times (2\hat{i}+3\hat{j}-\hat{k})

(2A+2B)=10i^+4j^+2k^(2A+2B)=10\hat{i}+4\hat{j}+2\hat{k}

A×(2A+2B)=∣i^j^k^3−121042∣A\times(2A+2B)=\begin{vmatrix} \hat{i} & \hat{j}&\hat{k} \\ 3& -1&2\\10&4&2 \end{vmatrix}

A×(2A+2B)=i^(−2−6)−j^(6−20)+k^(12+10)=−8i^+14j^+22k^A\times(2A+2B)=\hat{i}(-2-6)-\hat{j}(6-20)+\hat{k}(12+10)=-8\hat{i}+14\hat{j}+22\hat{k}

A.B=(3i^−j^+2k^).(2i^+3j^−k^)=6−6−2=−2A.B=(3\hat{i}-\hat{j}+2\hat{k}).(2\hat{i}+3\hat{j}-\hat{k})=6-6-2=-2


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