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Question: The area of the triangle whose vertices are A (1, –1, 2), B(2, 1, –1) and C (3, –1, 2) is -...

The area of the triangle whose vertices are A (1, –1, 2),

B(2, 1, –1) and C (3, –1, 2) is -

A

26

B

7137\sqrt{13}

C

13\sqrt{13}

D

8

Answer

13\sqrt{13}

Explanation

Solution

AB→=i^+2j^−3k^\overset{\rightarrow}{AB} = \widehat{i} + 2\widehat{j} - 3\widehat{k}

BC→=i^−2j^+3k^\overset{\rightarrow}{BC} = \widehat{i} - 2\widehat{j} + 3\widehat{k}

AB→×BC→=∣i^j^k^12−31−23∣=−6j^−4k^∣AB→×BC→∣=(−6)2+(−4)2=52\overset{\rightarrow}{AB} \times \overset{\rightarrow}{BC} = \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & - 3 \\ 1 & - 2 & 3 \end{matrix} \right| = - 6\widehat{j} - 4\widehat{k}|\overset{\rightarrow}{AB} \times \overset{\rightarrow}{BC}| = \sqrt{( - 6)^{2} + ( - 4)^{2}} = \sqrt{52} unit

\ Area of Δ=12∣AB→×BC→∣=13\Delta = \frac{1}{2}|\overset{\rightarrow}{AB} \times \overset{\rightarrow}{BC}| = \sqrt{13} unit