10. Area of the triangle enclosed by the vectors î +3j+2k,2i-j+k and - î+2 j+3k is
1)V117 units 2) V117/2 units 3) V117/2
4) none
Answers
Answer:
Given: Vectors : A =3i - 4j +2k and vector B = i + j - 2k as adjacent sides measured in metre
Area Of Triangle = 1/2 (A X B)
⇒ 1/2{}
= 1/2 (6i + 8j + 8k)
Magnitude = 1/2 ( √(6² + 8² + 8²) ) =√(164) / 2 ~ 6.403 Sq. units
∴ The Are of Traingle Formed ~ 6.403 Sq. units
The area of the triangle is √107 / 2 units²
Given: The vectors î + 3 j + 2 k , 2 i - j + k and - î + 2 j + 3 k.
To Find: Area of the triangle enclosed by the vectors î + 3 j + 2 k , 2 i - j + k and - î + 2 j + 3 k
Solution:
- We are given the vertices of a triangle. We shall consider a fixed point O, which is the origin. We will find the position vectors of each side of the triangle which will be required for calculation.
- The area of a triangle can be found by applying the formula,
Area = 1/2 | AB × AC | [ where, '×' denotes the cross product ]
- The vectors AB and AC are found by applying the position vectors.
AB = OB - OA and AC = OC - OA
where, OA, OB, and OC are position vectors.
Coming to the numerical,
Let the vertices of the triangle be A, B, and C, and let O be the fixed point.
∴ OA = î + 3 j + 2 k = ( 1, 3, 2 )
OB = 2 i - j + k = ( 2, -1, 1 )
OC = - î + 2 j + 3 k = ( -1, 2, 3 )
Now, we need to find the sides of the triangle,
∴ AB = OB - OA = ( 2, -1, 1 ) - ( 1, 3, 2 )
= ( 1, -4, -1 )
∴ AC = OC - OA = ( -1, 2, 3 ) - ( 1, 3, 2 )
= ( -2, -1, 1)
Now, we need to find the cross product, | AB × AC |
∴ AB × AC =
= i ( - 4 - 1 ) - j ( 1 - 2 ) + k ( - 1 - 8 )
= - 5 i + j - 9 k
∴ | AB × AC | = √ (( - 5 )² + ( 1 )² + ( - 9 )² )
= √107 units
Now, we already know that the area of a triangle can be found by applying the formula,
Area = 1/2 | AB × AC | [ where, '×' denotes the cross product ]
Applying this formula, we get;
Area = 1/2 × √107
= √107 / 2 units²
Hence, the area of the triangle is √107 / 2 units².
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