IIT JEE 1983 Maths - 'Adapted' - FIB to Single-Digit Integer Q4

Geometry Level 4

The area of the triangle whose vertices are A ( 1 , − 1 , 2 ) A(1, -1,2) , B ( 2 , 1 , − 1 ) B(2, 1,-1) , C ( 3 , − 1 , 2 ) C(3, -1,2) is A A sq. units. Find ⌊ A ⌋ \lfloor A\rfloor .

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The answer is 3.

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

A r e a o f t r i a n g l e A B C = 1 2 ∗ ∣ A B → × A C → ∣ . A B = B − A = ( 1 , 2 , − 3 ) , A C = C − A = ( 2 , 0 , 0 ) . A r e a = 1 2 ∗ ∣ x y z 1 2 − 3 2 0 0 ∣ = 1 2 ∗ ( 0 , 6 , 0 ) = 3. Area~of~ triangle~ ABC=\frac 1 2*|\overrightarrow{AB}\times\overrightarrow{AC}|.\\ AB=B - A=(1,2,-3),~~~~AC=C - A=(2,0,0).\\ Area~=\frac 1 2* \begin{vmatrix}x&y&z\\ 1&2&-3\\ 2&0&0\\ \end{vmatrix}= \frac 1 2*(0,6,0)\\ =3.

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