If A, B, C Are Three Non- Collinear Points with Position Vectors `Vec A, Vec B, Vec C`, Respectively, Then Show That The Length of the Perpendicular from Con Ab is `|(Vec a Xx Vec B)+(Vec B Xx Vec C) + (Vec B Xx Vec A)|/|(Vec B - Vec A)| - Mathematics

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If A, B, C are three non- collinear points with position vectors `vec a, vec b, vec c`, respectively, then show that the length of the perpendicular from Con AB is `|(vec a xx vec b)+(vec b xx vec c) + (vec b xx  vec a)|/|(vec b -  vec a)|`

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Solution

Let ABC be a triangle and let `vec a, vec b, vec c` be the position vectors of its vertices A, B, C respectively. Let CM be the perpendicular from C on AB. Then

Area of ΔABC = `1/2 (AB).CM = 1/2 |vec(AB)|(CM)`

Also Area of ΔABC = `1/2 |vec(AB) xx vec(AC)|`

Area of ΔABC = `1/2 |veca xx vecb + vecbxxvecc + vecc xx vec a|`

`:. 1/2|vec(AB)|(CM) = 1/2|vec a xx vec b + vec b xx vec c + vec c xx vec a|`

`=> CM = |vec a xx vec b + vec b xx vec c + vec c xx vec a|/|vec(AB)|`

`=> CM = |vec a xx vec b + vec b xx vec c + vec c xx vec a|/|vec b - vec a|`

  Is there an error in this question or solution?
2017-2018 (March) Set 1

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