Find the vector V with the given magnitude and the same direction as U.
Magnitude:
||v|| = 6
Direction:
U = <-3, 3>
Magnitude:
||v|| = 6
Direction:
U = <-3, 3>
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Notice that any vector that has the same direction as u must be parallel to u and must point in the same direction; in order for this to occur, the two vectors must be a positive scalar multiple of one another.
So, all vectors in the direction of u will take the form:
k<-3, 3> = <-3k, 3k>, for some k > 0.
Notice that k has to be positive; if k was negative, then k<-3, 3> will point in the direction OPPOSITE to that of u.
This vector has magnitude:
√[(-3k)^2 + (3k)^2] = √(9k^2 + 9k^2) = √(18k^2) = 3k√2.
We want the vector to have magnitude 6, so:
3k√2 = 6 ==> k = 6/(3√2) = 2/√2 = √2.
Therefore, the required vector is:
v = √2<-3, 3> = <-3√2, 3√2>.
I hope this helps!
So, all vectors in the direction of u will take the form:
k<-3, 3> = <-3k, 3k>, for some k > 0.
Notice that k has to be positive; if k was negative, then k<-3, 3> will point in the direction OPPOSITE to that of u.
This vector has magnitude:
√[(-3k)^2 + (3k)^2] = √(9k^2 + 9k^2) = √(18k^2) = 3k√2.
We want the vector to have magnitude 6, so:
3k√2 = 6 ==> k = 6/(3√2) = 2/√2 = √2.
Therefore, the required vector is:
v = √2<-3, 3> = <-3√2, 3√2>.
I hope this helps!
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To have the same direction, the coordinates need to be in the form <-a , a>, where a is a positive number.
Magnitude is sqrt(x^2+y^2), which in this case is sqrt((-a)^2 + a^2) = sqrt(2a^2); set that equal to 6.
square both sides, divide by two, take the square root: you should end up with 3 sqrt(2) for a.
So the vector V would be <-3sqrt2 , 3sqrt2>
Magnitude is sqrt(x^2+y^2), which in this case is sqrt((-a)^2 + a^2) = sqrt(2a^2); set that equal to 6.
square both sides, divide by two, take the square root: you should end up with 3 sqrt(2) for a.
So the vector V would be <-3sqrt2 , 3sqrt2>