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The angle between the vectors 3i+j+2k and 2i-2j+4k is
- a)cos-12/√7)
- b)sin-1(2/√7)
- c)cos-1(2/√5)
- d)sin-1(1/√7)
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The angle between the vectors 3i+j+2k and 2i-2j+4k isa)cos-12/√7...
Explanation:
Step 1: Find the dot product of the two vectors
- Given vectors:
Vector A = 3i + j + 2k
Vector B = 2i - 2j + 4k
- The dot product of two vectors A and B is given by: A.B = |A||B|cos(theta), where theta is the angle between the two vectors.
- Calculate the dot product of the given vectors:
A.B = (3*2) + (1*-2) + (2*4)
= 6 - 2 + 8
= 12
Step 2: Find the magnitudes of the vectors
- Magnitude of vector A: |A| = sqrt(3^2 + 1^2 + 2^2) = sqrt(14)
- Magnitude of vector B: |B| = sqrt(2^2 + (-2)^2 + 4^2) = sqrt(24) = 2sqrt(6)
Step 3: Find the cosine of the angle between the vectors
- Using the dot product formula: A.B = |A||B|cos(theta)
12 = sqrt(14) * 2sqrt(6) * cos(theta)
cos(theta) = 12 / (2sqrt(14)sqrt(6))
= 12 / (4sqrt(84))
= 3 / (sqrt(84))
- Simplifying the above expression gives cos(theta) = 1 / sqrt(7)
Step 4: Find the angle between the vectors
- The angle between the vectors can be found using the inverse trigonometric function:
theta = cos^(-1)(1 / sqrt(7))
Therefore, the angle between the vectors 3i+j+2k and 2i-2j+4k is sin^(-1)(2 / sqrt(7)).
Step 1: Find the dot product of the two vectors
- Given vectors:
Vector A = 3i + j + 2k
Vector B = 2i - 2j + 4k
- The dot product of two vectors A and B is given by: A.B = |A||B|cos(theta), where theta is the angle between the two vectors.
- Calculate the dot product of the given vectors:
A.B = (3*2) + (1*-2) + (2*4)
= 6 - 2 + 8
= 12
Step 2: Find the magnitudes of the vectors
- Magnitude of vector A: |A| = sqrt(3^2 + 1^2 + 2^2) = sqrt(14)
- Magnitude of vector B: |B| = sqrt(2^2 + (-2)^2 + 4^2) = sqrt(24) = 2sqrt(6)
Step 3: Find the cosine of the angle between the vectors
- Using the dot product formula: A.B = |A||B|cos(theta)
12 = sqrt(14) * 2sqrt(6) * cos(theta)
cos(theta) = 12 / (2sqrt(14)sqrt(6))
= 12 / (4sqrt(84))
= 3 / (sqrt(84))
- Simplifying the above expression gives cos(theta) = 1 / sqrt(7)
Step 4: Find the angle between the vectors
- The angle between the vectors can be found using the inverse trigonometric function:
theta = cos^(-1)(1 / sqrt(7))
Therefore, the angle between the vectors 3i+j+2k and 2i-2j+4k is sin^(-1)(2 / sqrt(7)).
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The angle between the vectors 3i+j+2k and 2i-2j+4k isa)cos-12/√7)b)sin-1(2/√7)c)cos-1(2/√5)d)sin-1(1/√7)Correct answer is option 'B'. Can you explain this answer?
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