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

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Answer 1 · 2018-06-06T06:01:08

$35.26°$

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

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

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

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

$color(white)(θ) = sin^-1(1/sqrt(3))$

$color(white)(θ) = 35.26°$