Can someone check my answer to this vector problem
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Can someone check my answer to this vector problem

[From: ] [author: ] [Date: 11-04-26] [Hit: ]
||V|| = √(9y² + y²) = 29 ==> 10y² = 29² ==> y = ±29/√(10).This gives two possible vectors V. or .You can write those with rationalized denominators if you prefer.......
Find a V whose magnitude of 29 and whose x component is 3 times the y component. Is the answer < 29 sqrt 15 over 4, 29/4> ?

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Maybe I'm reading your answer incorrectly. If that is <29√(15)/4, 29/4>, then the x component is not 3 times the y component. So that can't be right.

Put V = <3y,y>. Then

||V|| = √(9y² + y²) = 29 ==> 10y² = 29² ==> y = ±29/√(10).

This gives two possible vectors V.

<87/√(10), 29/√(10)> or <-87/√(10), -29/√(10)>.

You can write those with rationalized denominators if you prefer.
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