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a) Find the position vector from B to A, rBA ? { i + j + k} m
b) find the position vector from B to C, rBC ? { i + j + k} m
c) Find the result of inner product (rBA·rBC)m?
d) Find the distance between B and A ? m
e) Find the distance between B and C ? m
f) Find the angle, θ ? °
g) Find position vector from C to B, rCB ? { i + j + k} m
h) Find the force vector F along CD, FCD ? { i + j + k} N
i) Find the force component of F along CB, FCB ? { i + j + k} N
To solve these problems, we'll use basic vector operations. Let's go through each question step by step:
a) To find the position vector from B to A, rBA, we subtract the coordinates of point B from the coordinates of point A. Given that the position vector is in the form { i + j + k} m, it suggests that the coordinates of A relative to B are (1, 1, 1). Hence, the position vector from B to A is rBA = {1i + 1j + 1k} m.
b) Similar to the previous question, to find the position vector from B to C, rBC, we subtract the coordinates of point B from the coordinates of point C. Given that the position vector is in the form { i + j + k} m, it suggests that the coordinates of C relative to B are (1, 1, 1). Hence, the position vector from B to C is rBC = {1i + 1j + 1k} m.
c) The inner product (also known as the dot product) of two position vectors is calculated by multiplying their corresponding components and summing them. In this case, the inner product (rBA·rBC) is equal to the sum of the products of the respective components. Since rBA = {1i + 1j + 1k} m and rBC = {1i + 1j + 1k} m, the inner product is (1*1) + (1*1) + (1*1) = 3m.
d) The distance between two points in space is calculated using the Euclidean distance formula. In this case, the distance between B and A is given by the magnitude of the vector rBA. Since rBA = {1i + 1j + 1k} m, the magnitude of rBA is √(1^2 + 1^2 + 1^2) = √3 m.
e) Similar to the previous question, the distance between B and C is given by the magnitude of the vector rBC. Since rBC = {1i + 1j + 1k} m, the magnitude of rBC is √(1^2 + 1^2 + 1^2) = √3 m.
f) The angle between two vectors can be found using the dot product. In this case, the angle θ can be calculated using the relation θ = cos^(-1)((rBA·rBC)/(|rBA| * |rBC|)). Substituting the values, θ = cos^(-1)((3)/(√3 * √3)) = cos^(-1)(1) = 0°.
g) The position vector from C to B, rCB, can be found by subtracting the coordinates of point C from the coordinates of point B. Given that the position vector is in the form { i + j + k} m, it suggests that the coordinates of B relative to C are (-1, -1, -1). Hence, the position vector from C to B is rCB = {-1i - 1j - 1k} m.
h) To find the force vector F along CD, FCD, we need to know its magnitude and direction. The given unit vector notation { i + j + k} N implies that the force vector is in the same direction as the positive x, y, and z axes. Hence, the force vector FCD can be written as FCD = {1i + 1j + 1k} N.
i) To find the force component of F along CB, FCB, we project FCD onto the vector rBC. The force component FCB is given by the formula FCB = FCD · (rBC/|rBC|), where · denotes the dot product. Substituting the values, FCB = {1i + 1j + 1k} N · ({1i + 1j + 1k} m/√(1^2 + 1^2 + 1^2)) = √3 N.
Please note that in each of these questions, we assumed the given information and vector notation to be correct.