Question

The area enclosed by a triangle whose vertices Area(1,2,-3) B(0,0,0) C(2,4,7) is ?

Answer 1

Answer:

Step-by-step explanation:

Given, vertices of the triangle are $\tt{A(1,2,-3), B(0,0,0), C(2,4,7)}$

Now,

Let $\sf{\vec{a}}$ denotes the side BA of the triangle and $\sf{\vec{b}}$ denotes the side BC of the triangle.

So,

$\sf{\vec{a}=\hat{i}+2\hat{j}-3\hat{k}}$

And,

$\sf{\vec{b}=2\hat{i}+4\hat{j}+7\hat{k}}$

Now,

$\sf{Area\,\,\,of\,\,\,triangle\,\,\,,\triangle=\dfrac{1}{2}\,|\vec{a}\times\vec{b}|}$

$\sf{\implies\triangle=\dfrac{1}{2}\,\left|\left|\begin{array}{ccc}\sf{\hat{i}}&\sf{\hat{j}}&\sf{\hat{k}}\\1&2&-3\\2&4&7\end{array}\right|\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\,\left|\sf{\hat{i}}\left|\begin{array}{cc}\sf{2&-3\\4&7\end{array}\right|-\sf{\hat{j}}\left|\begin{array}{cc}\sf{1&-3\\2&7\end{array}\right|+\sf{\hat{k}}\left|\begin{array}{cc}\sf{1&2\\2&4\end{array}\right|\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\,\left|\sf{\hat{i}}(14+12)-\sf{\hat{j}}(7+6)+\sf{\hat{k}}(4-4)\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\,\left|26\sf{\hat{i}}-13\sf{\hat{j}}\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\,\times13\cdot\left|2\sf{\hat{i}}-\sf{\hat{j}}\right| }$

$\sf{\implies\triangle=\dfrac{1}{2}\,\cdot13\cdot\sqrt{4+1} }$

$\sf{\implies\triangle=\dfrac{1}{2}\,\cdot13\cdot\sqrt{5} }$

$\sf{\implies\triangle=\dfrac{13\sqrt{5}}{2} }$