Vertices of a triangle are given by i^+3j^+2k^,2i^−j^+k^ and −i^+2j^+3k^ , then the area of triangle is 1072n (in units). Find the value of n =

AB→=OB→−OA→=i^−4j^−k^AC→=OC→−OA→=−2i^−j^+k^|AB→×AC→|=i^j^k^1−4−1−2−11=−5i^+j^−9k^=107Area of triangle =12|AB→×AC→| =12107=1074∴n=2

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Vertices of a triangle are given by $\hat{i} + 3 \hat{j} + 2 \hat{k}, 2 \hat{i} - \hat{j} + \hat{k}$ and $- \hat{i} + 2 \hat{j} + 3 \hat{k}$ , then the area of triangle is $\sqrt{\frac{107}{2 n}} (in units).$ Find the value of n =

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Detailed Solution

$\vec{AB} = \vec{OB} - \vec{OA} = \hat{i} - 4 \hat{j} - \hat{k}$

$\vec{AC} = \vec{OC} - \vec{OA} = - 2 \hat{i} - \hat{j} + \hat{k}$

$| \vec{AB} \times \vec{AC} | = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 4 & - 1 \\ - 2 & - 1 & 1 \end{array}| = - 5 \hat{i} + \hat{j} - 9 \hat{k} = \sqrt{107}$

Area of triangle $= \frac{1}{2} | \vec{AB} \times \vec{AC} |$

$= \frac{1}{2} \sqrt{107} = \sqrt{\frac{107}{4}} ∴ n = 2$

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