Question

If $\widehat i - \widehat j + 2\widehat k$, $2\widehat i + \widehat j - \widehat k$ and $3\widehat i - \widehat j + 2\widehat k$ are position vectors of vertices of a triangle then its area is
(A)$26$
(B)$13$
(C)$2\sqrt {13} $
(D)$\sqrt {13} $

Answer 1

Hint: Find the vector form of any two adjacent sides of the triangle. Then use the formula of cross product to find the area of the triangle in vector form. Cross product is solved using determinants.

Complete step-by-step answer:
Let $\Delta ABC$ be a triangle whose vertices are given to us
Hence,$A = \widehat i - \widehat j + 2\widehat k$
$B = 2\widehat i + \widehat j - \widehat k$
$C = 3\widehat i - \widehat j + 2\widehat k$

We can write the vertices in terms of position vectors as
$\overline {OA} = \widehat i - \widehat j + 2\widehat k$
$\overline {OB} = 2\widehat i + \widehat j - \widehat k$
$\overline {OC} = 3\widehat i - \widehat j + 2\widehat k$
We know that, area of a triangle in vector form is given as
Area of triangle$ = \dfrac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right|$
Where, AB and BC are the adjacent sides of the triangle.
We can find AB as
$\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} $
$ = 2\widehat i + \widehat j - \widehat k - (\widehat i - \widehat j + 2\widehat k)$
$ = \left( {2 - 1} \right)\widehat i + \left( {1 + 1} \right)\widehat j + \left( { - 1 - 2} \right)\widehat k$
\[ \Rightarrow \overrightarrow {AB} = \widehat i + 2\widehat j - 3\widehat k\]
In the same way, we can find AC as
 $\overrightarrow {AC} = \overrightarrow {OC} - \overrightarrow {OA} $
$ = 3\widehat i - \widehat j + 2\widehat k - (\widehat i - \widehat j + 2\widehat k)$
$ = \left( {3 - 1} \right)\widehat i + \left( { - 1 + 1} \right)\widehat j + \left( {2 - 2} \right)\widehat k$
$ = 2\widehat i + 0\widehat j + 0\widehat k$
$ \Rightarrow \overrightarrow {AC} = 2\widehat i$
Cross product of two vectors is given by determinants as
$\overrightarrow {AB} \times \overrightarrow {AC} =
\begin{vmatrix} {\widehat i}&{\widehat j}&{\widehat k} \\
  {{a_1}}&{{a_2}}&{{a_3}} \\
  {{b_1}}&{{b_2}}&{{b_3}} \\
\end{vmatrix}$
Using this formula of determinant, we can write
$\overrightarrow {AB} \times \overrightarrow {AC}=
\begin{vmatrix} {\widehat i}&{\widehat j}&{\widehat k} \\
  1&2&{ - 3} \\
  2&0&0
\end{vmatrix}$
Expanding the determinant, we get
$\overrightarrow {AB} \times \overrightarrow {AC} = \widehat i\left( {0 - 0} \right) - \widehat j\left( {0 + 6} \right) + \widehat k\left( {0 - 4} \right)$
$ = - 6\widehat j - 4\widehat k$
$\overrightarrow {AB} \times \overrightarrow {AC} = - 6\widehat j - 4\widehat k$
Now using the formula of area of a triangle in vector form, we can write
$ \dfrac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \dfrac{1}{2}\left| { - 6\widehat j - 4\widehat k} \right|$
Since $\left| {a\widehat i + b\widehat j + c\widehat k} \right| = \sqrt {{a^2} + {b^2} + {c^2}} $
We can write
$\dfrac{1}{2}\left| {\overline {AB} \times \overline {BC} } \right| = \dfrac{1}{2}\sqrt {{{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} $
$ = \dfrac{1}{2}\sqrt {36 + 16} $
$ = \dfrac{1}{2}\sqrt {52} $
We would take $2$ inside the square root sign to simplify the expression
$ = \sqrt {\dfrac{{52}}{4}} $
$ = \sqrt {13} $
Hence the area of the triangle is $\sqrt {13} $ square units.
Therefore, from the above explanation, the correct answer is, option (D) $\sqrt {13} $
So, the correct answer is “Option C”.

Note: In this question, we got, $\overrightarrow {AB} \times \overrightarrow {AC} = - 6\widehat j - 4\widehat k$. Do not get confused with it or think that the answer is wrong as the area cannot be negative. Remember that a cross product is a vector product. So negative sign in this case only represents the direction of the product and not the magnitude of the product. Area is given by the magnitude of the product. That is why we take the magnitude of the cross product that we get. So, whether the cross product is negative or not, the magnitude will always be a positive quantity.