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Find the vector, not with determinants, but by using properties of cross products (ixj) x k?

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Final answer:

The vector resulting from (i × j) × k is zero, as per the properties of cross products in a three-dimensional space.

Step-by-step explanation:

The question asks us to find the vector resulting from the cross product (i × j) × k using the properties of cross products without relying on determinants. To solve this, we recall that the cross product of two unit vectors in a three-dimensional space follows some rules. The cross products i × i, j × j, and k × k all equal to zero since the cross product of a vector with itself is zero due to the sine of 0 degrees being zero.

Furthermore, when two unit vectors are crossed in a typical right-handed coordinate system, the result is the third unit vector, with the direction given by the right-hand rule. Referring to equations and principles related to cross products, the pattern is that crossing unit vectors in cyclic order (i, j, k) yields a positive unit vector, while crossing them out of order results in a negative sign. Hence, i × j = k, but j × i = -k.

Now, we apply this understanding to the vector at hand. Since i × j = k, we have k × k, as the first cross product already yields k. Following the rule that the cross product of any unit vector with itself is zero, we find that k × k = 0. Therefore, (i × j) × k simplifies to k × k, which finally equals zero.

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