Given vectors ( mathbf{A} = 3mathbf{i} - 2mathbf{j} + mathbf{k} ) and ( mathbf{B} = mathbf{i} + 2mathbf{j} - 2mathbf{k} ), find the angle between these two vectors.

Question

Accepted Answer

Step:1 ：First, we calculate the dot product of vectors ( mathbf{A} ) and ( mathbf{B} ). The dot product of two vectors ( mathbf{A} cdot mathbf{B} = |mathbf{A}||mathbf{B}|costheta ), where (|mathbf{A}|) and (|mathbf{B}|) are the magnitudes of ( mathbf{A} ) and ( mathbf{B} ) respectively, and (theta) is the angle between them. Thus, ( mathbf{A} cdot mathbf{B} = (3)(1) + (-2)(2) + (1)(-2) = -3 ).Step:2 ：Next, calculate the magnitudes of ( mathbf{A} ) and ( mathbf{B} ). The magnitude of a vector ( mathbf{A} = sqrt{(3)^2 + (-2)^2 + (1)^2} = sqrt{14} ), and the magnitude of vector ( mathbf{B} = sqrt{(1)^2 + (2)^2 + (-2)^2} = sqrt{9} = 3 ).Step:3 ：Substitute the dot product and the magnitudes of ( mathbf{A} ) and ( mathbf{B} ) into the formula for the dot product: ( costheta = frac{mathbf{A} cdot mathbf{B}}{|mathbf{A}||mathbf{B}|} = frac{-3}{sqrt{14} cdot 3} = -frac{1}{sqrt{14}} ).Step:4 ：Finally, find the angle (theta) by taking the inverse cosine (or arccos) of (-frac{1}{sqrt{14}}).

Answer

Step: 1 ：First, we calculate the dot product of vectors ( mathbf{A} ) and ( mathbf{B} ). The dot product of two vectors ( mathbf{A} cdot mathbf{B} = |mathbf{A}||mathbf{B}|costheta ), where (|mathbf{A}|) and (|mathbf{B}|) are the magnitudes of ( mathbf{A} ) and ( mathbf{B} ) respectively, and (theta) is the angle between them. Thus, ( mathbf{A} cdot mathbf{B} = (3)(1) + (-2)(2) + (1)(-2) = -3 ).Step: 2 ：Next, calculate the magnitudes of ( mathbf{A} ) and ( mathbf{B} ). The magnitude of a vector ( mathbf{A} = sqrt{(3)^2 + (-2)^2 + (1)^2} = sqrt{14} ), and the magnitude of vector ( mathbf{B} = sqrt{(1)^2 + (2)^2 + (-2)^2} = sqrt{9} = 3 ).Step: 3 ：Substitute the dot product and the magnitudes of ( mathbf{A} ) and ( mathbf{B} ) into the formula for the dot product: ( costheta = frac{mathbf{A} cdot mathbf{B}}{|mathbf{A}||mathbf{B}|} = frac{-3}{sqrt{14} cdot 3} = -frac{1}{sqrt{14}} ).Step: 4 ：Finally, find the angle (theta) by taking the inverse cosine (or arccos) of (-frac{1}{sqrt{14}}).

Given vectors $A = 3 i - 2 j + k$ and $B = i + 2 j - 2 k$ , find the angle between these two vectors.

Answer

Popular Problems

Given vectors A=3i−2j+k and B=i+2j−2k, find the angle between these two vectors.

Answer

Popular Problems

Given vectors $A = 3 i - 2 j + k$ and $B = i + 2 j - 2 k$ , find the angle between these two vectors.