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Question
A triangle is formed by joining the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). Find the direction cosines of the medians
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Solution
`vec"OA" = hat"i", vec"OB" = hat"j", vec"OC" = hat"k"`
D is the midpoit of BC
∴ `vec"OD" = (vec"OB" + "OC")/2`
= `(hat"j" + hat"k")/2`
E is the midpoint of AC
`vec"OE" = (hat"i" + hat"k")/2`
F is the midpoint of AB
∴ `vec"OF" = (hat"i" + hat"j")/2`
Now the medians are `vec"AD", vec"BE"` and `vec"CF"`
(i) `vec"AD" =vec"OD" - vec"OA" = (hat"j" +hat"k")/2 - hat"i"`
= `-hat"i" + hat"j"/2 + hat"k"/2`
`|vec"AD"| = sqrt(1 + 1/4 + 1/4)`
= `sqrt(1 + 1/2)`
= `sqrt(3)/sqrt(2)`
and d.c's of `vec"AD" = ((-1)/(sqrt(3)/sqrt(2)), (1/2)/(sqrt(3)/sqrt(2)), (1/2)/(sqrt(3)/sqrt(2)))`
= `(- sqrt(2)/sqrt(3), sqrt(2)/(2sqrt(3)), sqrt(2)/(2sqrt(3)))`
= `(- sqrt(2)/sqrt(3), 1/(sqrt(2)sqrt(3)), 1/(sqrt(2)/sqrt(3)))`
= `(- sqrt(2)/sqrt(3), 1/sqrt(6), 1/sqrt(6))`
`["Now" sqrt(2)/sqrt(3) = sqrt(2)/sqrt(3) xx sqrt(2)/sqrt(2) = 2/sqrt(6)] = (- 2/sqrt(6), 1/sqrt(6), 1/sqrt(6))`
(ii) `vec"BE" = vec"OE" - vec"OB"`
= `(hat"i" + hat"k")/2 -hat"j"`
= `hat"i"/2 - hat"j" + hat"k"/2`
`|vec"BE"| = sqrt(1/4 + 1 + 1/4)`
= `sqrt(3)/sqrt(2)`
d.c's of `vec"BE" = ((1/2)/(sqrt(3)/sqrt(2)), (-1)/(sqrt(3)/sqrt(2)), (1/2)/(sqrt(3)/sqrt(2))) = (1/sqrt(6), (-2)/sqrt(6), 1/sqrt(6))`
(iii) `vec"CE" = vec"OF" - vec"OC"`
= `(hat"i" + hat"j")/2 - hat"k"`
= `hat"i"/2 + hat"j"/2 - hat"k"`
`|vec"CF"| = sqrt(1/4 + 1/4 + 1)`
= `sqrt(3)/sqrt(2)`
d.c's of `vec"CF" = ((1/2)/(sqrt(3)/sqrt(2)), (1/2)/(sqrt(3)/sqrt(2)), (-1)/(sqrt(3)/sqrt(2)))`
= `(1/sqrt(6), 1/sqrt(6), (-2)/sqrt(6))`
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