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Find the area of the triangle whose vertices are A(3, – 1, 2), B(1, – 1, – 3) and C(4, – 3, 1) - Mathematics

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प्रश्न

Find the area of the triangle whose vertices are A(3, – 1, 2), B(1, – 1, – 3) and C(4, – 3, 1)

बेरीज

उत्तर

The given vertices of the triangle ABC are A(3, – 1, 2), B(1, – 1, – 3) and C(4, – 3, 1)

`vec"OA" = 3hat"i" - hat"j" + 2hat"k"`

`vec"OB" = hat"i" - hat"j" - 3hat"k"`

`vec"OC" = 4hat"i" - 3hat"j" + hat"k"`

`vec"AB" = vec"OB" - vec"OA"`

= `(hat"i" - hat"j" - 3hat"k") - (3hat"i" - hat"j" + 2hat"k")`

= `hat"i" - hat"j" - 3hat"k" - 3hat"i" + hat"j" - 2hat"k"`

`vec"AB" = -2hat"i" - 5hat"k"`

`vec"AC" = vec"OC" - vec"OA"`

= `(4hat"i" - 3hat"j" + hat"k") - (3hat"i" - hat"j" + 2hat"k")`

= `4hat"i" - 3hat"j" + hat"k" - 3hat"i" + hat"j" - 2hat"k"`

`vec"AC" = hat"i" - 2hat"j" - hat"k"`

`vec"AB" xx vec"AC" = |(hat"i", hat"j", hat"k"),(-2, 0, -5),(1, -2, -1)|`

= `hat"i"(0 - 10) - hat"j"(2 + 5) + hat"k"(4 - 0)`

= `-10hat"i" - 7hat"j" + 4hat"k"`

`|vec"AB" xx vec"AC"| = |-10hat"i" - 7hat"j" + 4hat"k"|`

= `sqrt((-10)^2 + (-7)^2 + 4^2`

= `sqrt(100 + 49 + 16)`

= `sqrt(165)`

Area of the triangle ABC = `1/2 |vec"AB" xx vec"AC"|`

= `1/2 xx sqrt(165)`

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पाठ 8: Vector Algebra - Exercise 8.4 [पृष्ठ ८०]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 8 Vector Algebra
Exercise 8.4 | Q 6 | पृष्ठ ८०

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