Find the vector, not with determinants, but by using properties of cross products. (i+j)×(i−j)

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Brittany Patton · Accepted Answer

Calculations:  Apply the distribution property of the cross product, we get  (i+j)×(i−j)=i×i−i×j+j×i−j×j  And since the angle between the cross product i×i and j×j is zero, and since sin⁡(0∘)=0 then from equation the cross product for any two vectors having the same direction is also zero, substituting we get  =0−i×j+j×i−0  =−i×j+j×i  Applying, the commutation property of the cross product, we get  =−i×j−i×j  Applying equation, we get  =−2(|a||b|sin⁡(θ)n)  And since the magnitude of the unit direction i and j is 1, and the angle between both vectors is 90∘, thus sin⁡θ=1 therefor we have  =−2(1)(1)(1)n  i×j=−2n  And, the direction of the unit vector n is orthogonal to i×j, is determined using the right hand rule, where curling the fingers from the i to j the thumb would be pointing toward the positive direction of k, therefor  n=k  And, thus the final answer is  =−2k  Result: -2k

Jeffrey Jordon · Accepted Answer

Now we find (i+j)×(i−j)  (i+j)×(i−j)=i×(i−j)+j×(i−j)  (i+j)×(i−j)=i×i−i×j+j×i−j×j  (i+j)×(i−j)=0−k+(−k)−0  (i+j)×(i−j)=−2k

Find the vector, not with determinants, but by using properties of cross products. (i+j)\times(i-j)

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