Using the definition of the scalar product, find the angles between the following pairs of vectors.
A = - 4 i + 2 j + 2 k and B = - 3 j - 3 k
I understand to do [(A).(B)]/|A|*|B| = cos theta; but I'm getting 0 for |A| and don't know how to access it from there.
never mind, found the mistake I was making.
I AM NOT ABLE TO FIND ANGLE
To find the angles between two vectors using the scalar product, you need to calculate the magnitudes of the vectors involved. The magnitude of a vector is calculated using the formula:
|A| = √(A_x^2 + A_y^2 + A_z^2)
Let's calculate the magnitudes of vectors A and B:
For vector A:
|A| = √((-4)^2 + 2^2 + 2^2)
= √(16 + 4 + 4)
= √24
For vector B:
|B| = √(0^2 + (-3)^2 + (-3)^2)
= √(0 + 9 + 9)
= √18
Now that we have the magnitudes of vectors A and B, we can calculate the scalar product:
(A·B) = A_x * B_x + A_y * B_y + A_z * B_z
Substituting the given values:
(A·B) = (-4)(0) + (2)(-3) + (2)(-3)
= 0 + (-6) + (-6)
= -12
Now, we can find the angle between the vectors using the formula you mentioned:
cos(theta) = (A·B) / (|A| * |B|)
Substituting the calculated values:
cos(theta) = -12 / (√24 * √18)
Now, simplify the expression:
cos(theta) = -12 / (√(24 * 18))
= -12 / √(432)
= -12 / (2√108)
= -12 / (2 * 6√3)
= -1 / (6√3)
To determine the angle, take the inverse cosine (cos^-1) of this value:
theta = cos^-1(-1 / (6√3))
Therefore, the angle between vectors A and B is theta = cos^-1(-1 / (6√3)).