Find angle between unit vectors I+j+k cap And I+j cap by cross product?

1 Answer
Jun 6, 2018

#35.26°#

Explanation:

#vec"A" = hati + hatj + hatk#
#vec"B" = hati + hatj#

Magnitude of vectors are

#|vec"A"| = sqrt(1^2 + 1^2 + 1^2) = sqrt(3)#
#|vec"B"| = sqrt(1^2 + 1^2) = sqrt(2)#

Cross product of two vectors are

#vec"A" × vec"B" = (hati + hatj + hatk) × (hati + hatj)#

#color(white)(vec"A" × vec"B") = (hati × hati) + (hati × hatj) + (hatj × hati) + (hatj × hatj) + (hatk × hati) + (hatk × hatj)#

#color(white)(vec"A" × vec"B") = 0 + hatk + (-hatk) + 0 + hatj + (-hati)#

#color(white)(vec"A" × vec"B") = hatj - hati#

Magnitude of cross product is

#|vec"A" × vec"B"| = sqrt(1^2 + 1^2) = sqrt(2)#

Angle between them is

#theta = sin^-1[(|vec"A" × vec"B"|)/(|vec"A"||vec"B"|)]#

#color(white)(θ) = sin^-1[cancel(sqrt(2))/(sqrt(3) · cancel(sqrt(2)))]#

#color(white)(θ) = sin^-1(1/sqrt(3))#

#color(white)(θ) = 35.26°#