Find the angle between the vectors A = (3, -2, 1) and B = (1, 1, 1).
Practice Questions
1 question
Q1
Find the angle between the vectors A = (3, -2, 1) and B = (1, 1, 1).
60°
45°
90°
30°
cos(θ) = (A · B) / (|A| |B|). A · B = 3*1 + (-2)*1 + 1*1 = 2. |A| = √(3^2 + (-2)^2 + 1^2) = √14, |B| = √3. θ = cos^(-1)(2/(√14 * √3)).
Questions & Step-by-step Solutions
1 item
Q
Q: Find the angle between the vectors A = (3, -2, 1) and B = (1, 1, 1).
Solution: cos(θ) = (A · B) / (|A| |B|). A · B = 3*1 + (-2)*1 + 1*1 = 2. |A| = √(3^2 + (-2)^2 + 1^2) = √14, |B| = √3. θ = cos^(-1)(2/(√14 * √3)).
Steps: 10
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. This is done by multiplying the corresponding components of A and B and then adding them together: A · B = 3*1 + (-2)*1 + 1*1.
Step 3: Perform the calculations for the dot product: A · B = 3 - 2 + 1 = 2.
Step 4: Calculate the magnitude of vector A, |A|. Use the formula |A| = √(3^2 + (-2)^2 + 1^2).
Step 5: Perform the calculations for |A|: |A| = √(9 + 4 + 1) = √14.
Step 6: Calculate the magnitude of vector B, |B|. Use the formula |B| = √(1^2 + 1^2 + 1^2).
Step 7: Perform the calculations for |B|: |B| = √(1 + 1 + 1) = √3.
Step 8: Use the formula for the cosine of the angle θ: cos(θ) = (A · B) / (|A| |B|).
Step 9: Substitute the values into the formula: cos(θ) = 2 / (√14 * √3).
Step 10: To find the angle θ, use the inverse cosine function: θ = cos^(-1)(2 / (√14 * √3)).
