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Exercise 8.2 · Q4

Q.A triangle is formed by joining the points (1,0,0),(0,1,0)(1,0,0),(0,1,0) and (0,0,1)(0,0,1). Find the direction cosines of the medians.

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Step 1. Let A(1,0,0),B(0,1,0),C(0,0,1)A(1,0,0),B(0,1,0),C(0,0,1).

Step 2 (median from A). Midpoint of BCBC is D=(0,12,12)D=\left(0,\tfrac12,\tfrac12\right). AD⃗=D−A=(−1,12,12)\vec{AD}=D-A=\left(-1,\tfrac12,\tfrac12\right), or direction ratios (−2,1,1)(-2,1,1) after doubling. r=4+1+1=6r=\sqrt{4+1+1}=\sqrt6, so DCs =(−26,16,16)=\left(\tfrac{-2}{\sqrt6},\tfrac1{\sqrt6},\tfrac1{\sqrt6}\right).

Step 3 (median from B). Midpoint of ACAC is E=(12,0,12)E=\left(\tfrac12,0,\tfrac12\right). BE⃗=E−B=(12,−1,12)→\vec{BE}=E-B=\left(\tfrac12,-1,\tfrac12\right)\to DRs (1,−2,1)(1,-2,1), DCs =(16,−26,16)=\left(\tfrac1{\sqrt6},\tfrac{-2}{\sqrt6},\tfrac1{\sqrt6}\right). …

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