6. When A=31-2j, B=i+2j-k. Then unit vector along (A-B) would be
(a) 1/√21(i-4j+k) (b) 1/√21 (2i-4j + k)
(c) 1/√3(2i-4j+ k) (d) 1/√3 (2î - 4j + k)
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6. When A=31-2j, B=i+2j-k. Then unit vector along (A-B) would be
(a) 1/√21(i-4j+k) (b) 1/√21 (2i-4j + k)
(c) 1/√3(2i-4j+ k) (d) 1/√3 (2î - 4j + k)
tell me in process
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Explanation:
To find the unit vector along the vector (A - B), you need to calculate (A - B) first, then normalize it to a unit vector. Here are the calculations:Given: A = 31 - 2j B = i + 2j - kNow, find A - B:A - B = (31 - 2j) - (i + 2j - k)= 31 - 2j - i - 2j + k= (31 - i) - 4j + kNow, calculate the magnitude of (A - B):|(A - B)| = √[(31 - i)² + (-4j)² + k²] = √[(31 - i)² + 16 + 1] (since j² = -1)= √[(31 - i)² + 17]Now, to find the unit vector along (A - B), you need to divide (A - B) by its magnitude:Unit vector = (A - B) / |(A - B)|= [(31 - i) - 4j + k] / √[(31 - i)² + 17]None of the provided answer options matches the unit vector that would result from these calculations. However, you can simplify the unit vector expression, and it should be in the form of (x/√21)i + (y/√21)j + (z/√21)k, where x, y, and z are the respective components of the unit vector
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