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victus00 [196]
3 years ago
8

What are the correct answers?

Mathematics
1 answer:
DochEvi [55]3 years ago
6 0

Step-by-step Answer:

Situations with dot-products.

dot-product of vectors A<xa,ya>, B<xb,yb>

is defined as A.B=xa*xb + ya+yb

Note that the result is a scalar even though A and B are both vectors.

Q1 dot product is positive

Q2 dot product is negative

The dot product is also equal to the length of the projection of the unit vector of one onto the other, multiplied by the magnitudes.  This implies that the sign will be positive if the angle between the vectors is acute, and negative if the angle between them is obtuse.  That makes the first two proposed answers (A, B) correct.

Q3 dot product equals 0

Since the projection of any vector onto another perpendicular to it is zero, so the dot product of two perpendicular vectors always equal zero.  That makes the third proposed answer (E) correct.

Q4 dot product equals the product of the vectors' magnitudes.

The earlier statement

"The dot product is also equal to the length of the projection of the unit vector of one onto the other, multiplied by the magnitudes."

makes the fourth statement true if the unit vector projected onto the other remains a unit vector, i.e. when the two vectors are parallel (D)

Q5 dot product equals opposite of the product of the vectors.

To have the dot product's absolute value equal the product of the magnitudes, the vectors must be parallel.  If the dot product is positive, then the vectors are parallel and pointing in the same direction, as in Q4.

If the dot product is negative, meaning that is the opposite of the product of the magnitudes (themselves are always positive or zero), then the vectors are parallel, but in the opposing directions.

There is no correct answer for this case, which is definitely NOT "not possible".  The closest one is (D) vectors are parallel, but answer is incomplete.

Q6 dot product is greater than the product of the vectors' magnitudes.

I do not see a possible case for this to happen.


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Hi there! :)

<u>Answer:</u>

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