Find the angle between the vectors A = (3, -2, 1) and B = (1, 1, 1) if A · B = |A||B|cos(θ).

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Q: Find the angle between the vectors A = (3, -2, 1) and B = (1, 1, 1) if A · B = |A||B|cos(θ).

Solution: A · B = 3*1 + (-2)*1 + 1*1 = 3 - 2 + 1 = 2. |A| = √(3^2 + (-2)^2 + 1^2) = √14, |B| = √3. cos(θ) = 2/(√14 * √3). θ = 60°.

Steps: 13

Step 1: Identify the vectors A and B. A = (3, -2, 1) and B = (1, 1, 1).
Step 2: Calculate the dot product A · B using the formula A · B = A1*B1 + A2*B2 + A3*B3.
Step 3: Substitute the values: A · B = 3*1 + (-2)*1 + 1*1.
Step 4: Simplify the calculation: A · B = 3 - 2 + 1 = 2.
Step 5: Calculate the magnitude of vector A using the formula |A| = √(A1^2 + A2^2 + A3^2).
Step 6: Substitute the values: |A| = √(3^2 + (-2)^2 + 1^2) = √(9 + 4 + 1) = √14.
Step 7: Calculate the magnitude of vector B using the formula |B| = √(B1^2 + B2^2 + B3^2).
Step 8: Substitute the values: |B| = √(1^2 + 1^2 + 1^2) = √(1 + 1 + 1) = √3.
Step 9: Use the formula A · B = |A||B|cos(θ) to find cos(θ).
Step 10: Substitute the values: 2 = √14 * √3 * cos(θ).
Step 11: Solve for cos(θ): cos(θ) = 2 / (√14 * √3).
Step 12: Use the inverse cosine function to find θ: θ = cos⁻¹(2 / (√14 * √3)).
Step 13: Calculate θ to find the angle in degrees, which is approximately 60°.