P=(-2,2) Q=(3,4) S=(2,-8)
PS-3PQ
Find the component form and magnitude of the vector.
PS-3PQ
= (-10,4) - 3(2,5)
= (-10,4) + (-6,-15)
= (-16,-11)
magnitude is √((-16)^2 + (-11)^2)
= √377
thank You
welcome
I dont understand how you got the answer. could you pls explain?
To find the component form of a vector, we subtract the corresponding coordinates of the starting point from the corresponding coordinates of the ending point.
Given:
P = (-2,2)
Q = (3,4)
To find the vector PQ, we subtract the coordinates of point P from the coordinates of point Q:
PQ = (3-(-2), 4-2)
= (5, 2)
Next, to find the vector PS, we subtract the coordinates of point P from the coordinates of point S:
PS = (2-(-2), -8-2)
= (4, -10)
To find the vector PS - 3PQ, we subtract three times the coordinates of the vector PQ from the coordinates of the vector PS:
PS - 3PQ = (4, -10) - 3(5, 2)
= (4, -10) - (15, 6)
= (4-15, -10-6)
= (-11, -16)
So, the component form of the vector PS - 3PQ is (-11, -16).
To find the magnitude of a vector, we use the Pythagorean theorem:
The magnitude ||v|| of a vector v = (x, y) is given by ||v|| = √(x^2 + y^2).
For the vector PS - 3PQ = (-11, -16), the magnitude is:
||PS - 3PQ|| = √((-11)^2 + (-16)^2)
= √(121 + 256)
= √377
≈ 19.43
So, the magnitude of the vector PS - 3PQ is approximately 19.43.