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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the area of the triangle with vertices $A(1, 1, 2), B(2, 3, 5)$ and $C(1, 5, 5)$.

$\begin{array}{1 1}(A) \large\frac{\sqrt{61}}{2} sq.units \\ (B) \large\frac{\sqrt{48}}{2} sq.units \\ (C) \sqrt{48} sq.units \\(D) \sqrt{61} sq.units\end{array} $

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  • Area of a triangle with sides $AB,BC$ and $AC$ is given by $\large\frac{1}{2}|$$\overrightarrow{AB}\times \overrightarrow{BC}|$
Step 1:
Given vertices of the triangle are $A(1,1,2),B(2,3,5)$ and $C(1,5,5)$
Let the position vectors be $\overrightarrow a,\overrightarrow b$ and $\overrightarrow c$
Hence $\overrightarrow a=\hat i+\hat j+2\hat k$
$\qquad\overrightarrow b=2\hat i+3\hat j+5\hat k$
$\qquad\overrightarrow c=\hat i+5\hat j+5\hat k$
Now let us determine $\overrightarrow{AB}=\overrightarrow b-\overrightarrow a$
$\qquad\qquad\qquad\qquad\quad=(2\hat i+3\hat j+5\hat k)-(\hat i+\hat j+2\hat k)$
$\qquad\qquad\qquad\qquad\quad=\hat i+2\hat j+3\hat k$
Step 2:
Next let us determine $\overrightarrow{AC}=\overrightarrow c-\overrightarrow a$
$\qquad\qquad\qquad\qquad\quad=(\hat i+5\hat j+5\hat k)-(\hat i+\hat j+2\hat k)$
$\qquad\qquad\qquad\qquad\quad=4\hat j+3\hat k$
Step 3:
The area of the triangle is given by $\large\frac{1}{2}$$|\overrightarrow {AB}\times \overrightarrow {AC}|$
$\overrightarrow {AB}\times \overrightarrow {AC}=\begin{vmatrix}\hat i & \hat j& \hat k\\1 & 2 & 3\\0 & 4 & 3\end{vmatrix}$
$\qquad\qquad=\hat i(6-12)-\hat j(3-0)+\hat k(4-0)$
$\qquad\qquad=-6\hat i-3\hat j+4\hat k$
Step 4:
$|\overrightarrow {AB}\times \overrightarrow {AC}|=\sqrt{(-6)^2+(-3)^2+4^2}$
$\qquad\qquad=\sqrt{36+9+16}$
$\qquad\qquad=\sqrt{61}$
Step 5:
The area of the triangle = $\large\frac{1}{2}$$|\overrightarrow {AB}\times \overrightarrow {AC}|$
We know $|\overrightarrow {AB}\times \overrightarrow {AC}|=\sqrt{61}$
Therefore the area of the triangle=$\large\frac{\sqrt {61}}{2}$sq.units.
answered May 22, 2013 by sreemathi.v
 

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