Using vectors, find the area of the triangle ABC, whose vertices are $A (1, 2, 3), B (2, - 1, 4)$ and $C (4, 5, - 1)$ .

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Solution

Given vertices of $Δ A B C$ are A(1, 2, 3), B(2, -1, 4) and C(4, 5, -1)
So, $- - \to A B = - - \to O B - - - \to O A = (2 \to i - \to j + 4 \to k) - (\to i + 2 \to j + 3 \to k) = \to i - 3 \to j + \to k$ and
$- - \to A C = - - \to O C - - - \to O A = (4 \to i + 5 \to j - \to k - (\to i + 2 \to j + 3 \to k)) = 3 \to i + 3 \to j - 4 \to k$
Now $- - \to A B \times - - \to A C = ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to i & \to j & \to k 1 & - 3 & 1 3 & 3 & - 4 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 9 \to i + 7 - ----- \to j + 12 \to k$
$\Rightarrow ∣ ∣ ∣ - - \to A B \times - - \to A C ∣ ∣ ∣ = \sqrt{9^{2} + 7^{2} + 12^{2}} = \sqrt{274}$
$a r (A B C) = \frac{1}{2} ∣ ∣ ∣ - - \to A B \times - - \to A C ∣ ∣ ∣ = \frac{1}{2} \sqrt{274}$ Sq.units.

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Using vectors, find the area of the $△$ ABC with vertices A(1, 2, 3), B (2, -1, 4) and C(4, 5, -1).

A (1, 2, 3), B (2, - 1, 1)

C (1, 2, - 4)

A (1, 2, 3), B (2, - 1, 6), C (3, 2, - 3)

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