Show that  a.b = (1/4) |a+b|^2 - (1/4) |a-b|^2

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justaguide | College Teacher | (Level 2) Distinguished Educator

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Let us take the two vectors a and b.

Their dot product is |a|*|b|*cos D, where D is the angle between the vectors.

Let the vector a make an angle of A with the horizontal and vector b make an angle of B with the horizontal.

(1/4)*|a + b|^2 - (1/4)*|a - b|^2

=> (1/4)[|a + b|^2 - (1/4)*|a - b|^2]

=> (1/4)[[sqrt((|a|*sin A + |b|*sin B)^2 + (|a|*cos A + |b|*cos B)^2)]^2 - [sqrt((|a|*sin A - |b|*sin B)^2 + (|a|*cos A - |b|*cos B)^2)]^2 ]]

=> (1/4)[[sqrt(|a|^2*(sin A)^2 + |b|^2*(sin B)^2 + 2|a|*|b|*sin A * sin B+ |a|^2*(cos A)^2 + |b|^2*(cos B)^2 + 2|a|*|b|*cos A * cos B]^2 -  [sqrt((|a|^2*(sin A)^2 + |b|^2*(sin B)^2 - 2|a|*|b|*sin A * sin B+ |a|^2*(cos A)^2 + |b|^2*(cos B)^2 - 2|a|*|b|*cos A * cosB]^2]

simplify using (cos x)^2 + (sin x)^2 = 1

=> (1/4)[[(|a|^2 + 2|a|*|b|*sin A * sin B + 2|a|*|b|*cos A * cos B] -  [(|a|^2 + |b|^2 - 2|a|*|b|*sin A * sin B - 2|a|*|b|*cos A * cos B]]

=> (1/4)[4|a|*|b|*sin A * sin B + 4|a|*|b|*cos A * cos B]

=> [|a|*|b|*sin A * sin B + |a|*|b|*cos A * cos B]

use cos (A - B) = cos A * cos B + sin A * sin B

=> |a|*|b|*cos ( A - B)

=> |a|*|b|*cos D

which is a.b

Therefore we prove that a.b = (1/4)*|a + b|^2 - (1/4)*|a - b|^2

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