Question

Find the area of the triangle whose two sides are represented by vector
( 2i + 3j+k)m and (i + ſ + k)m.
​

Answer 1

The area of the triangle is $\dfrac{\sqrt6}{2}$ m².

Explanation:

Let the sides be represented by A and B.

So, vector A is given as $\overrightarrow A=(2i + 3j+k)\ m$

Vector B is given as $\overrightarrow B=(i +j+ k)\ m$

Area of the triangle formed by the sides A and B is given as:

Area = $\frac{1}{2}(|\overrightarrow A\times \overrightarrow B|)$

So, let us first determine the magnitude of the cross product of vectors A and B.

The cross-product is given as:

$\overrightarrow A\times \overrightarrow B\\=(2i + 3j+k)\times (i +j+ k)\\=(2i\times i)+(2i\times j)+(2i\times k) +(3j\times i)+(3j\times j)+(3j\times k)+(k\times i)+(k\times j)+(k\times k)\\=0+2k-2j-3k+0+3i+j-i+0\\=2i-j-k$

Now, magnitude of the cross product is given as:

$|\overrightarrow A\times \overrightarrow B|=\sqrt{2^2+(-1)^2+(-1)^2}\\|\overrightarrow A\times \overrightarrow B|=\sqrt{4+1+1}=\sqrt{6}$

Therefore, the area of the triangle is:

$Area=\frac{1}{2}\times \sqrt{6}\\Area=\frac{\sqrt6}{2}\ m^2$