Question

Define the scalar product ā, b of two vectors ā and b
someone help me with this ASAP
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Answer 1

Scalar product is defined as the product of magnitudes of two vectors and the cosine of the angle between the vectors.

unit vector u,v,w along U,V,W is :

• u = U/|U| =  (2i + 3j -6k)/7

• v = V/|V| =  (6i + 2j + 3k)/7

• w = W/|W| = ( 3i - 6j - 2k)/7

• Also if a = x1i +y1j + z1k

           and b = x2i + y2j+ z2k,

           then a.b = x1.x2 + y1.y2 + z1.z2

• The scalar product of two vectors a and b, is also known as dot product of a and a unit vector, b is the projection/shadow of vector a on a unit vector b.

• For eg: If two vectors are perpendicular to each other, and we keep a light from top of vector a.  Then no shadow will fall on b. That is its projection is zero.

• ie. a.b = 0 , if a and b is perpendicular.

Keeping this in mind,

We need to show that U,V,W are mutually perpendicular,

• U = 2i + 3j -6k, V = 6i + 2j + 3k, W = 3i - 6j - 2k

• U.V = 2x6 + 3x2 + -6x3 = 12+6-18=0

• V.W = 6x3 + 2x-6 +3x-2 = 18-12-6 = 0

• W.X = 3x2 -6x3 -2x-6 = 6 -18 +12 = 0

This implies U,V,W are mutually perpendicular.

|U| = $\sqrt{2^{2}+3^{2} + 6^{2} } = \sqrt{49} = 7$

unit vector u,v,w along U,V,W is :

• u = U/|U| =  (2i + 3j -6k)/7

• v = V/|V| =  (6i + 2j + 3k)/7

• w = W/|W| = ( 3i - 6j - 2k)/7