Find the magnitude and direction angle of the vector v.
v= 8(cos135 degrees i + sin135 degrees J)
I think that the answer is v= 5 and Q = 135. Am I correct?
not quite
v= 8(cos135 degrees i + sin135 degrees J)
= 8(-1/√2, 1/√2)
the magnitude of (-1/√2, 1/√2) is 1
so the magnitude of 8(-1/√2, 1/√2) is 8
Your direction is correct
Thank You!!
No, you are not correct. To find the magnitude and direction angle of the vector v, we can use the formula:
Magnitude (r) = √(x^2 + y^2)
Direction angle (θ) = tan^(-1)(y/x)
Given that v = 8(cos135 degrees i + sin135 degrees j), we can extract the x and y components:
x = 8(cos135 degrees) = -8(cos45 degrees) = -8(1/sqrt(2)) = -8/sqrt(2) = -4sqrt(2)
y = 8(sin135 degrees) = 8(sin45 degrees) = 8(1/sqrt(2)) = 8/sqrt(2) = 4sqrt(2)
Now, we can substitute these values into the formula to find the magnitude:
Magnitude (r) = √((-4sqrt(2))^2 + (4sqrt(2))^2) = √(32 + 32) = √64 = 8
To find the direction angle, we can substitute the x and y values into the formula:
Direction angle (θ) = tan^(-1)((4sqrt(2))/(-4sqrt(2))) = tan^(-1)(-1) = -45 degrees
Therefore, the magnitude of vector v is 8 and the direction angle is -45 degrees.
To find the magnitude and direction angle of the vector v, you are correct in using the formula:
v = magnitude * (cos(angle) i + sin(angle) j)
Given that v = 8(cos135 degrees i + sin135 degrees j), we can see that the magnitude is 8 and the angle is 135 degrees.
Thus, the magnitude of vector v is 8, and the direction angle is 135 degrees. Therefore, your answer is correct.