the scalar product of vector A = 3i+2j and vector B is 10? find vector B
If B = ai+bj then
3a+2b = 10
I guess B=2i+2j will work
To find vector B, we can use the formula for the scalar product:
A · B = |A| |B| cos(theta)
Given that A = 3i + 2j and A · B = 10, let's find |A| first:
|A| = √(3^2 + 2^2) = √(9 + 4) = √13
Now, rearrange the formula to solve for |B|:
10 = |A| |B| cos(theta)
10 = √13 |B| cos(theta)
Since the cosine of an angle can never exceed 1, we have cos(theta) = 1. Therefore, the equation becomes:
10 = √13 |B|
Now, solve for |B|:
|B| = 10 / √13
So, the magnitude of vector B is 10 / √13.
To find the direction of vector B, we can use unit vector notation. Divide the vector A by its magnitude to obtain the unit vector uA:
uA = A / |A| = (3i + 2j) / √13
Now, the direction of vector B is parallel to the direction of vector A. Therefore, B can be represented as follows:
B = |B| uA
B = (10 / √13) uA
Hence, vector B is (10 / √13) times the unit vector uA.
To find vector B, we can use the formula for the scalar product (also known as the dot product) between two vectors:
A · B = |A| |B| cosθ
Given that the scalar product of vector A and vector B is 10, we can write the equation as:
3*|B| * cosθ + 2*|B| * cosθ = 10
To simplify further, let's assume that the magnitude of vector B is |B| = a, and let's solve for the value of cosθ.
5a * cosθ = 10
Dividing both sides by 5a, we get:
cosθ = 10 / (5a) = 2 / a
Now, to find vector B, we need to determine the direction of vector B, which is given by the angle θ. Since the cosine function is positive in the first and fourth quadrants, we know that θ is either in the first quadrant (0° to 90°) or the fourth quadrant (270° to 360°).
To find the actual value of θ, we can use the inverse cosine function (also known as arccosine):
θ = arccos(2 / a)
Now that we have the magnitude and angle of vector B, we can express it as a vector by using the polar representation:
B = a(cosθ i + sinθ j)
Substituting the value of θ, we get:
B = a(cos(arccos(2 / a)) i + sin(arccos(2 / a)) j)
Simplifying further:
B = a(2 / a i + √(1 - (2/a)^2) j)
B = 2i + √(a^2 - 4) j
So, the vector B is given by 2i + √(a^2 - 4) j, where a is a positive value representing the magnitude of vector B.