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Let a and b be two non-collinear unit vectors. if
u = a–(a.b) b and v = a×b, then |v| isa)|u|b)|u|+ |u.a|c)|u|+ |u.b|d)|u|+ u.(a b)Correct answer is option 'B,C'. Can you explain this answer?
Verified Answer
Let a and b be two non-collinear unit vectors. if ... moreu = a–(a.b) ...
Correct option: (b, c)
Explanation:

Let θ be the angle between vector a and b. Since, vector a and b are non-collinear vectors, then θ ≠0 and θ ≠ π.





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Most Upvoted Answer
Let a and b be two non-collinear unit vectors. if ... moreu = a–(a.b) ...
Solution:

Given,
u = a – (a.b) b
v = a x b

To find |v|, we need to find the magnitude of the vector v.

Finding |v|:

v = a x b
|v| = |a x b|
|v| = |a| |b| sinθ
Since a and b are unit vectors, |a| = |b| = 1
|v| = sinθ

Finding θ:

θ is the angle between vectors a and b.
cosθ = a.b
θ = cos⁻¹ (a.b)

Finding u:

u = a – (a.b) b
u = a – (a.b) (b/|b|)
u = a – (a.b/|b|) b

Finding |u|:

|u| = √[a² + (a.b/|b|)² - 2(a.b/|b|) (a.b/|b|)]
|u| = √[1 + (a.b)²/|b|² - 2(a.b)²/|b|²]
|u| = √[(|b|² + (a.b)² - 2(a.b)²)/|b|²]
|u| = √[(|b|² - (a.b)²)/|b|²]
|u| = √[(1 - (a.b)²)]

Option B:

|v| = sinθ
θ = cos⁻¹ (a.b)
|v| = sin(cos⁻¹ (a.b))
|v| = √(1 - (a.b)²)

Option C:

|u| = √[(1 - (a.b)²)]

Therefore, the correct answer is B and C.
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Let a and b be two non-collinear unit vectors. if ... moreu = a–(a.b) b and v = a×b, then |v| isa)|u|b)|u|+ |u.a|c)|u|+ |u.b|d)|u|+ u.(a b)Correct answer is option 'B,C'. Can you explain this answer?
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