IF A= 2x+3y and B=-x+y,then the angle between the resultant A+B and the positive directio of x-axis is ;.....
Ax=2, Bx= - 1 => (A+B)x=2-1=1
Ay=3, By= 1 => (A+B)y=3+1=4
tanα=(A+B)y/(A+B)x =4/1=4,
α =arctan4=76º
Ax=2, Bx= - 1 => (A+B)x=2-1=1
Ay=3, By= 1 => (A+B)y=3+1=4
tanα=(A+B)y/(A+B)x =4/1=4,
α =arctan4=76º
To find the angle between the resultant A + B and the positive direction of the x-axis, you'll need to compute the dot product of the resultant vector and the unit vector in the positive direction of the x-axis. Let's go step by step:
1. Find the resultant vector (A + B):
A = 2x + 3y
B = -x + y
Add the two vectors together:
A + B = (2x + 3y) + (-x + y)
= x + 4y
So, the resultant vector is R = x + 4y.
2. Compute the dot product:
To compute the dot product, multiply the x-components and the y-components of the two vectors and add them together.
Let's assume the angle between the resultant vector and the positive direction of the x-axis is θ. Also, let the unit vector in the positive direction of the x-axis be i.
The dot product is given by:
R · i = |R| * |i| * cos(θ)
Since the unit vector i has a magnitude of 1, we can simplify it to:
R · i = |R| * cos(θ)
We need to find the magnitude |R|. To find this, we take the square root of the sum of the squares of the x and y components of R:
|R| = sqrt((1^2) + (4^2))
= sqrt(1 + 16)
= sqrt(17)
The dot product becomes:
R · i = sqrt(17) * cos(θ)
3. Solve for θ:
Rearrange the equation to isolate cos(θ):
cos(θ) = (R · i) / |R|
Substitute the values we found:
cos(θ) = (sqrt(17) * cos(θ)) / sqrt(17)
The square root of 17 cancels out, leaving:
cos(θ) = cos(θ)
Since the cosine function is equal for θ and (2π - θ), we have:
θ = 2π - θ
Simplifying, we get:
2θ = 2π
θ = π
Therefore, the angle between the resultant vector (A + B) and the positive direction of the x-axis is π (180 degrees).