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tierann46
29.11.2020 •
Mathematics
Im not sure how to answer this question.
u= 5i-2j+3k v= -2i+3k
Find a vector that is in the direction of the vector v , whose magnitude is twice that of vector u
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Ответ:
9514 1404 393
(-4√494)/13i +(6√494)/13k ≈ -6.8388i +0j +10.2585k
Step-by-step explanation:
To answer this question, you need to know two things:
1) the direction of vector v
2) the magnitude of vector u
__
direction of v
A direction is specified by a "unit vector", one with the proper ratio of components, and a magnitude of 1. It is found from a given vector by dividing that vector by its magnitude.
The unit vector in the v direction is ...
v/|v| = (-2i +3k)/(√((-2)² +3²) = (-2i +3k)/√13
__
magnitude of u
The magnitude of vector u is ...
|u| = √(5² +(-2)² +3²) = √38
__
Then the desired vector is ...
(2|u|)(v/|v|) = 2√38(-2i+3k)/√13 = (-4√494)/13i +(6√494)/13k
Additional comment
We have chosen to "rationalize the denominator" by writing √(38/13) as (√494)/13.
Ответ:
Hope i will be the brainliest answer