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Find the angles of a triangle whose verticesare `A(0,-1,-2),B(3,1,4)`and `C`(5,7,1). |
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Answer» Here, `vec(AB) = vecB - vecA = 3hati+2hatj+6hatk` `vec(BC) = 2hati+6hatj-3hatk` `vec(AC) = 5hati+8hatj+3hatk` `|vec(AB)| = sqrt(3^2+2^2+6^2) = 7` `|vec(BC)| = sqrt(32^2+6^2+(-3)^2) = 7` `|vec(AC)| = sqrt(5^2+8^2+3^2) = 7sqrt2` Now, `vec(AB)*vec(BC) = 6+12-18 = 0` It means, `AB` and `BC` are perpendicular. `:./_ABC = 90^@` Also, as `|vec(AB)| = |vec(BC)| ` `:. /_ACB = /_BAC` As, `/_ABC = 90^@` `:. /_ACB = /_BAC = 1/2*(180-90)^@ = 45^@.` |
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