Vector
A
has a magnitude of 29 units and points in the positive y-direction. When vector
B
is added to
A,
the resultant vector
A + B
points in the negative y-direction with a magnitude of 15 units. Find the magnitude and direction of
B?
A + B = -15
29 + B = -15
B = -15-29 =-44 Units and points in negative Y direction.
Well, it seems like Vector A is really trying to make a positive impact, pointing in the positive y-direction! But then Vector B comes along and ruins its day, leading to a resultant vector pointing in the negative y-direction. Classic case of vector interference, am I right?
Now let's get down to business. Since Vector A has a magnitude of 29 units and Vector A + B has a magnitude of 15 units, we can assume that the magnitude of Vector B is 29 - 15 = 14 units. So, Vector B is not as large as Vector A, but it still has some significance.
As for the direction of Vector B, since the resultant vector points in the negative y-direction, it means that Vector B must be pointing in the opposite direction to Vector A. In other words, Vector B is pointing in the negative y-direction.
To summarize, the magnitude of Vector B is 14 units, and it points in the negative y-direction. Just remember, even though Vector B may be a bit of a troublemaker, it still plays an important role in the greater scheme of vectors!
To solve this problem, let's break it down step-by-step:
Step 1: Express vector A in terms of its components.
Since vector A points in the positive y-direction, we can express it as A = (0, 29), where the x-component is 0 and the y-component is 29.
Step 2: Express vector B in terms of its components.
Let's assume the x-component of vector B is Bx and the y-component is By. Therefore, vector B can be expressed as B = (Bx, By).
Step 3: Determine the resultant vector A + B.
The resultant vector A + B is the vector that points from the tail of vector A to the head of vector B. Given that the resultant vector points in the negative y-direction and has a magnitude of 15 units, we can express it as A + B = (0, -15).
Step 4: Combine the components of vectors A, B, and A + B.
When we add the x-components of vectors A and B, we get: 0 + Bx = 0, which implies Bx = 0.
When we add the y-components of vectors A and B, we get: 29 + By = -15. Solving for By, we have By = -15 - 29 = -44.
Step 5: Determine the magnitude of vector B.
The magnitude of vector B can be calculated using the Pythagorean theorem: |B| = sqrt(Bx^2 + By^2). In this case, Bx = 0, so |B| = sqrt(0^2 + (-44)^2) = sqrt(1936) = 44 units.
Step 6: Determine the direction of vector B.
The direction of vector B can be found using the arctangent function: θ = arctan(By / Bx). In this case, Bx = 0, so the direction of vector B is θ = arctan(-44 / 0). Since this is undefined, the direction of vector B is vertical and points directly downward.
Therefore, the magnitude of vector B is 44 units, and its direction is downward in the negative y-direction.
To find the magnitude and direction of vector B, we need to break down the information given in the problem.
We know that vector A has a magnitude of 29 units and points in the positive y-direction. This means that A is a vector in the coordinate plane with coordinates (0, 29).
We also know that when vector B is added to A, the resultant vector A + B points in the negative y-direction with a magnitude of 15 units. This means that the resulting vector A + B is a vector in the coordinate plane with coordinates (0, -15).
To find vector B, we can use the following equation:
A + B = (0, -15)
Adding A to both sides of the equation:
A + A + B = A + (0, -15)
Simplifying:
2A + B = (0, -15)
Since vector A has a magnitude of 29 units and points in the positive y-direction, its coordinates are (0, 29). Plugging in the values:
(2 * 0, 2 * 29) + B = (0, -15)
(0, 58) + B = (0, -15)
Now, to isolate vector B, we need to subtract (0, 58) from both sides of the equation:
B = (0, -15) - (0, 58)
Subtracting the coordinates:
B = (0 - 0, -15 - 58)
Simplifying:
B = (0, -73)
Therefore, vector B has a magnitude of 73 units (since |-73| = 73) and points in the negative y-direction.