Physics, asked by goswaminagendra749, 12 hours ago

Find The Area Of a Triangle P(1 3,2) Q(2,-1,1) And R(-1,2,3).



Need Step By step calculation..​

Answers

Answered by pulakmath007
5

SOLUTION

TO DETERMINE

The Area Of a Triangle P(1 3,2) Q(2,-1,1) And R(-1,2,3).

EVALUATION

Here the given points are P(1 3,2) Q(2,-1,1) And R(-1,2,3)

Let O be the origin

Then

 \vec{OP} =  \hat{i} + 3 \hat{j} + 2 \hat{k}

\vec{OQ} =  2\hat{i}  -  \hat{j} +  \hat{k}

\vec{OR} = -   \hat{i} + 2 \hat{j} + 3 \hat{k}

Thus we get

\vec{PQ} =  \hat{i}  - 4 \hat{j}  -  \hat{k}

\vec{PR} =  - 2 \hat{i}  -  \hat{j} +  \hat{k}

Now

\vec{PQ}  \times \vec{PR}

 = \displaystyle\begin{vmatrix}  \hat{i} &  \hat{j} &  \hat{j}\\ 1 &  - 4 &   - 1 \\  - 2 &  - 1 &  1 \end{vmatrix}

 =  \hat{i}( - 4 - 1)  -   \hat{j}(1 - 2)   + \hat{j}( - 1 - 8)

 =   - 5\hat{i}  +   \hat{j}    - 9\hat{j}

Thus we get

 |\vec{PQ}  \times \vec{PR} |

 \sf{ =  \sqrt{ {( - 5)}^{2}  +  {(1)}^{2}  +  {( - 9)}^{2} } }

 \sf{ =  \sqrt{ 25 + 1 + 81} }

 \sf{ =  \sqrt{107} }

Hence the required area of the triangle PQR

 \displaystyle \:  =  \frac{1}{2}  |\vec{PQ}  \times \vec{PR} |  \:  \:  \: sq.unit

 \displaystyle \:  =  \frac{1}{2}   \sqrt{107}   \:  \:  \: sq.unit

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