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Question

Mathematics Question on angle between two lines

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

A

454 \sqrt{5} s units

B

232 \sqrt{3} s units

C

13 \sqrt{13} s units

D

15 \sqrt{15} s units

Answer

13 \sqrt{13} s units

Explanation

Solution

Area of ΔABC=12BA×BC\Delta\: ABC = \frac{1}{2} | \overrightarrow{BA} \times \overrightarrow{BC} |
BA=i^2j^+3k^\overrightarrow{BA} = - \hat{i} - 2 \hat{j} + 3 \hat{k}
BC=i^2j^+3k^\overrightarrow{BC} = \hat{i} - 2 \hat{j} + 3 \hat{k}
\therefore Area = 12i^j^k^ 1\-23 123 \frac{1}{2} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ -1& \- 2&3\\\ 1& -2 & 3 \end{vmatrix}
=166j^+4k^= \frac{1}{6} | 6 \, \hat{j} + 4 \hat{k} |
=3j^+2k^=9+4=13= |3 \hat{j} + 2 \hat{k} | = \sqrt{9 + 4} = \sqrt{13} s units