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

The area of a triangle whose vertices are A(1,−1,2)A(1, - 1,2), B(2,1,−1)B(2,1, - 1) and C(3,−1,2)C(3, - 1,2) is

A

13

B

13\sqrt{13}

C

6

D

6\sqrt{6}

Answer

13\sqrt{13}

Explanation

Solution

AB→=(2i+j−k)−(i−j+2k)=i+2j−3k\overset{\rightarrow}{AB} = (2\mathbf{i} + \mathbf{j} - \mathbf{k}) - (\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}, AC→=(3i−j+2k)−(i−j+2k)=2i\overset{\rightarrow}{AC} = (3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) - (\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = 2\mathbf{i}Area of triangle

ABC= 12∣AB→×AC→∣\frac{1}{2}|\overset{\rightarrow}{AB} \times \overset{\rightarrow}{AC}|

= 12∣(i+2j−3k)×2i∣\frac{1}{2}|(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) \times 2\mathbf{i}| = 12∣−4k−6j∣=∣−3j−2k∣=13\frac{1}{2}| - 4\mathbf{k} - 6\mathbf{j}| = | - 3\mathbf{j} - 2\mathbf{k}| = \sqrt{13}