How do I find the angle between the planes 5(x+1) + 3(y+2) + 2z = 0 and x + 3(y-1) + 2(z+4) = 0?

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Answer 1 · 2016-09-04T19:47:13

#~~38.69^@#

Let the given planes be #pi_1 and pi_2#. Let #vecn_1 and vec n_2#

be their normals.

Recall that for a plane #: ax+by+cz+d=0#, its normal #vecn=(a,b,c)#.

Clearly, #vec n_1=(5,3,2), and, n_2=(1,3,2)#

the #/_alpha# btwn. #pi_1 and pi_2# is, by defn.,

# alpha=arc cos{|vecn_1*vecn_2|/(||vecn_1||||vecn_2||)}#.

#=arc cos{|5*1+3*3+2*2|/((sqrt(25+9+4)sqrt(1+9+4))}}#

#=arc cos(18/(sqrt(2*19*2*7)))#

#=arc cos(9/sqrt(133))#

#=arc cos(9/11.53)#

#~~arc cos (0.7806)#

#~~38.69^@#