Find the magnitude of the dot product of the two vectors B=6i+j+2k, D=-1+3j. What is the angle between B and D
To find the magnitude of the dot product of vectors B and D, we first need to calculate the dot product. The dot product of two vectors A and B is given by the formula:
A · B = A_x * B_x + A_y * B_y + A_z * B_z
where A_x, A_y, A_z are the components of vector A, and B_x, B_y, B_z are the components of vector B.
In this case, vector B is given as B = 6i + j + 2k, and vector D is given as D = -1 + 3j.
Thus, the dot product of B and D is:
B · D = (6 * -1) + (1 * 3) + (2 * 0) = -6 + 3 + 0 = -3
The magnitude of the dot product is the absolute value of the dot product:
|B · D| = |-3| = 3
Now, to find the angle between B and D, we can use the formula:
cos θ = (B · D) / (|B| * |D|)
where θ is the angle between the vectors, |B| is the magnitude of vector B, and |D| is the magnitude of vector D.
The magnitude of vector B is:
|B| = √(6^2 + 1^2 + 2^2) = √(36 + 1 + 4) = √41
The magnitude of vector D is:
|D| = √((-1)^2 + 3^2) = √(1 + 9) = √10
Now we can substitute the values into the formula for cos θ:
cos θ = (-3) / (√41 * √10)
To find the angle θ, we can take the inverse cosine (cos^-1) of the result:
θ = cos^-1((-3) / (√41 * √10))
Hence, the magnitude of the dot product of vectors B and D is 3, and the angle between B and D is θ, given by the equation above.