find the angles of the triangle ABC where A=(2,3,-4) B=(1,1,-2) and C=(-1,7,1)

Question

find the angles of the triangle ABC where A=(2,3,-4) B=(1,1,-2) and
C=(-1,7,1)​

Sanjay Kumar Manjhi · Accepted Answer

Dear Student,Coordinate of vertices of △ABC are; A(2, 3, -4)A⇀=2i⇀+3j⇀-4k⇀similarly, B(1, 1, -2)B⇀=i⇀+j⇀-2k⇀and C(-1, 7, 1)C⇀=-i⇀+7j⇀+k⇀Now, AC⇀=(C⇀-A⇀)=-3i⇀+4j⇀+5k⇀and, CA⇀=-AC⇀=3i⇀-4j⇀-5k⇀AB⇀=(B⇀-A⇀)=-i⇀-2j⇀+2k⇀and,BA⇀=- AB⇀=i⇀+2j⇀-2k⇀BC⇀=-2i⇀+6j⇀+3k⇀and,CB⇀=- BC⇀=2i⇀-6j⇀-3k⇀Now,∠A is angle between vector AB⇀ and vector AC⇀;So, AB⇀·AC⇀=|AB⇀|×|AC⇀|×cosθ=cosθ=AB⇀·AC⇀|AB⇀|×|AC⇀|Now, AB⇀·AC⇀=(-i⇀-2j⇀+2k⇀)·(-3i⇀+4j⇀+5k⇀)=3-8+10=5|AB⇀|=-12+(-2)2+22=1+4+4=5and, |AC⇀|=-32+42+52=9+16+25=52So, cosθ=55×52=110=θ=cos-1(110)=θ=71
56 degreeSo, ∠A =cos-1(110)=71
56°Now, lets find out ∠B which is angle between vector BA⇀ and vector BC⇀; BA⇀=i⇀+2j⇀-2k⇀ and BC⇀=-2i⇀+6j⇀+3k⇀Now, BA⇀· BC⇀=-2+12-6=4|BA⇀|=12+22+(-2)2=9=3|BC⇀|=-22+62+32=49=7cosθ=BA⇀· BC⇀|BA⇀|×|BC⇀|=43×7=cosθ=421=θ=cos-1(421)=79
01°So, ∠B =cos-1(421)=79
01°Now, lets find out ∠C which is angle between vector CA⇀ and vector CB⇀;CA⇀=3i⇀-4j⇀-5k⇀ and CB⇀=2i⇀-6j⇀-3k⇀Now, CA⇀·CB⇀=6+24+15=45|CA⇀|=32+(-4)2+(-5)2=50=52| CB⇀|=22+(-6)2+(-3)2=49=7cosθ=CA⇀·CB⇀|CA⇀|×| CB⇀|=4552×7=cosθ=972=θ=cos-1(972)=24
61°So, ∠C=cos-1(972)=24 61°Regards

find the angles of the triangle ABC where A=(2,3,-4) B=(1,1,-2) and C=(-1,7,1)​