Question

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

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asked 2021-06-09
Find the vector, not with determinants, but by using properties of cross products. \((i+j)\times(i-j)\)

Answers (1)

2021-06-10
Calculations:
Apply the distribution property of the cross product, we get
\((i+j)\times(i-j)=i\times i-i\times j+j\times i-j\times j\)
And since the angle between the cross product \(i\times i\) and \(j\times j\) is zero, and since \(\sin(0^\circ)=0\) then from equation the cross product for any two vectors having the same direction is also zero, substituting we get
\(=0-i\times j+j\times i-0\)
\(=-i\times j+j\times i\)
Applying, the commutation property of the cross product, we get
\(=-i\times j-i\times j\)
Applying equation, we get
\(=-2(|a||b|\sin(\theta)n)\)
And since the magnitude of the unit direction i and j is 1, and the angle between both vectors is \(90^\circ\), thus \(\sin\theta=1\) therefor we have
\(=-2(1)(1)(1)n\)
\(i\times j=-2n\)
And, the direction of the unit vector n is orthogonal to \(i\times j\), 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
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