Question

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



Need Step By step calculation..​

Answer 1

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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