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A=i^-j^ 2k^ & b= 3i^-3j^ 6k^ are (1)parallel (2) perpendicular (3) inclined to an acute angle (4) inclined yo an obtuse angle.??? Reply plz?
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A=i^-j^ 2k^ & b= 3i^-3j^ 6k^ are (1)parallel (2) perpendicular (3) inc...
Assuming you meant A=(i^-j^)2k^,

We can simplify this expression by first evaluating the term inside the parentheses:

i^-j^ = 1/i^j^

Using the rules of exponents, we can rewrite this as:

i^-j^ = 1/(i^j)

Now we can substitute this back into the original expression:

A = (1/(i^j))^2k^

Using another rule of exponents, we can simplify this further:

A = 1/(i^2jk)

Therefore, the simplified form of A=(i^-j^)2k^ is A = 1/(i^2jk).
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A=i^-j^ 2k^ & b= 3i^-3j^ 6k^ are (1)parallel (2) perpendicular (3) inc...
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A=i^-j^ 2k^ & b= 3i^-3j^ 6k^ are (1)parallel (2) perpendicular (3) inclined to an acute angle (4) inclined yo an obtuse angle.??? Reply plz?
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