Let a and b be two non-collinear unit vectors. If u=a-(a ·b) b and v=a ×b, then |v| is (A) |u| (

Let a and b be two non-collinear unit vectors. If u=a-(a ·b)...

Key Concepts

Unit Vectors

Unit vectors are vectors with a magnitude of one. They serve as the building blocks in vector analysis and are used to indicate direction. When working with unit vectors, many formulas simplify because the scale (magnitude) factors become one, which is useful in both dot and cross product computations.

Dot Product

The dot product, or scalar product, of two vectors is a measure of their directional alignment and is computed as the product of their magnitudes and the cosine of the angle between them. This operation is widely used for projecting one vector onto another and for determining the angle between vectors.

Cross Product

The cross product, or vector product, of two vectors results in a third vector that is perpendicular to the plane containing the original vectors. The magnitude of the cross product is given by the product of the magnitudes of the two vectors and the sine of the angle between them, capturing the area of the parallelogram spanned by the vectors.

Vector Projection and Orthogonal Decomposition

Vector projection involves decomposing a vector into components parallel and perpendicular to another vector. The expression a - (a · b) b represents the component of vector a that is orthogonal to vector b, isolating the portion of a that does not align with b. This concept is fundamental in understanding how vectors interact in different directions within a plane.

Transcript

00:01 Hello friends, so in this problem we have been given let a and b be to non -collinear unit vectors.

00:09 If u equal to a minus a dot b into b and b equal to a cross b then magnitude of v is asked.

00:15 So let us proceed with the problem.

00:20 So now let theta be angle between unit vectors a and b.

00:26 As a and b are non -colinear we can infer the that theta is not equal to 0 and theta is not equal to pi.

00:36 We have a dot b equal to magnitude of a into magnitude of b into cos theta.

00:43 That is equal to cost theta.

00:47 So now since these are unit vectors it will be cost theta.

00:51 So now magnitude of u squared equal to magnitude of a minus a dot b into b squared, which is equal to magnitude of a minus cost theta into b whole square so this will be equal to magnitude of a square plus cos square theta magnitude of b square square minus two cos theta a dot b so now we have this equal to 1 plus cos square theta minus 2 cosquare theta which is equal to 1 minus cosquare theta which is equal to sine squared theta so now or here we have also magnitude of v squared equal to magnitude of a cross b squared that is equal to magnitude of a squared magnitude of b squared through sine square theta so that implies magnitude of v square equal to magnitude of u square that implies magnitude of v uh u equal to magnitude of v so also we have u dot a equal to a minus a dot b into b dot a which is equal to which is equal to a dot a minus a dot b into b dot a so this will be equal to magnitude of a squared minus cos square theta.

02:44 That is equal to 1 minus cosquare theta, which is again sine square theta.

02:50 So now moving forward, we have therefore magnitude of u plus magnitude of u...

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