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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 - Mathematics

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प्रश्न

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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उत्तर


`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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Vector Algebra - Exercise 8.2 [पृष्ठ ६८]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 8 Vector Algebra
Exercise 8.2 | Q 4 | पृष्ठ ६८

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