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Question

If r =xi^+yj^+zk^, then write the value of r ×i^2.

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Solution

Given:r=xi^+yj^+zk^Now,i=i^+0j^+0k^r×i=i^j^k^xyz100 =0 i^+zj^-yk^r×i=z2+y2r×i2=z2+y2

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