Question

find the angle between the vector a=3i+4j-5k and b=2I-3j-7k
​

Answer 1

Answer:

we know angle between any two vector is theta so here if we modified the formula a bar. b bar =abcostheta we get,

cos theta =a bar. b bar /ab so,

cos theta =(3i+4j-5k).( 2i-3j-7k) /ab hence,

cos theta=(3) (2) +(4) (-3) +(-5) (-7) /ab hene a=√3^2+4^2+(-5) ^2=√9+16+25=√50=2√5 like such

b=√2^2+(-3)^2+(-7)^2=√4+9+49=√62 now put this values of a and b in formula we get, cos theta=(3) (2) +(4) (-3) +(-5) (-7)/2√5×√62 hence cos theta =6-12+35/2√310=29/2√310 hence as cos theta =29/2√310 therefore theta =cos^-1(29/2√310) hence angle between those two vectors is cos^-1(29/2√310)