Question

Calculate the area of the triangle determined by the two vectors:
\[\mathop A\limits^ \to = \mathop {3i}\limits^ \wedge + \mathop {4j}\limits^ \wedge \] and \[\mathop B\limits^ \to = \mathop { - 3i}\limits^ \wedge + \mathop {7j}\limits^ \wedge \].

Answer 1

Hint:
As we know, the area of a triangle is the total region that is enclosed by three sides. To calculate the area of the triangle of the two vectors, apply formula by taking the two vectors given. Hence formula to calculate the area is:
Area of a triangle = \[\dfrac{1}{2}|\mathop A\limits^ \to \times \mathop B\limits^ \to |\]

Complete step by step solution:
To find the area of the triangle let us apply the vector formula for finding the area.
Area of a triangle = \[\dfrac{1}{2}|\mathop A\limits^ \to \times \mathop B\limits^ \to |\] ……………………… 1
Where, A and B are the vectors given as
\[\mathop A\limits^ \to = \mathop {3i}\limits^ \wedge + \mathop {4j}\limits^ \wedge \] and
\[\mathop B\limits^ \to = \mathop { - 3i}\limits^ \wedge + \mathop {7j}\limits^ \wedge \].
Substitute the vectors A and B in equation 1
Area of a triangle = \[\dfrac{1}{2}|\mathop A\limits^ \to \times \mathop B\limits^ \to |\]
= \[\dfrac{1}{2}|\left( {\mathop {3i}\limits^ \wedge + \mathop {4j}\limits^ \wedge } \right) \times \left( {\mathop { - 3i}\limits^ \wedge + \mathop {7j}\limits^ \wedge } \right)|\]
Now let us calculate the value of \[|\mathop A\limits^ \to \times \mathop B\limits^ \to |\] as A and B are in determinant so let us find it.
\[|\mathop A\limits^ \to \times \mathop B\limits^ \to |\] = \[|\left( {\begin{array}{*{20}{c}}
  {\mathop i\limits^ \wedge }&{\mathop j\limits^ \wedge }&{\mathop k\limits^ \wedge } \\
  3&4&0 \\
  { - 3}&7&0
\end{array}} \right)|\]
= \[\mathop i\limits^ \wedge \] \[\left( {\begin{array}{*{20}{c}}
  4&0 \\
  7&0
\end{array}} \right)\]-\[\mathop j\limits^ \wedge \] \[\left( {\begin{array}{*{20}{c}}
  3&0 \\
  { - 3}&0
\end{array}} \right)\] +\[\mathop k\limits^ \wedge \]\[\left( {\begin{array}{*{20}{c}}
  3&4 \\
  { - 3}&7
\end{array}} \right)\]
After simplifying we get
= \[\mathop i\limits^ \wedge \left( {4 \cdot 0 - 7 \cdot 0} \right) - \mathop j\limits^ \wedge \left( {3 \cdot 0 + 3 \cdot 0} \right) + \mathop k\limits^ \wedge \left( {3 \cdot 7 - \left( { - 3} \right) \cdot 4} \right)\]
= \[\mathop i\limits^ \wedge |0| - \mathop j\limits^ \wedge |0| + \mathop k\limits^ \wedge |21 + 12|\]
= \[|21\mathop k\limits^ \wedge + 12\mathop k\limits^ \wedge |\]

Hence the value of \[|\mathop A\limits^ \to \times \mathop B\limits^ \to |\] is
= \[|33\mathop k\limits^ \wedge |\]
To get the Area of triangle now let us substitute the value of \[|\mathop A\limits^ \to \times \mathop B\limits^ \to |\] in equation 1.
Area of a triangle = \[\dfrac{1}{2}|\mathop A\limits^ \to \times \mathop B\limits^ \to |\]
Area of a triangle = \[\dfrac{1}{2}|33k|\]
Therefore,
Area of a triangle is 16.5 square units.

Additional information:
In vector theory, vectors are visualized as directed line segments whose lengths are their magnitudes. Normally when we try to find out the area of a triangle, we usually find out the value by the formula of Heron’s Formula.
It is applicable to all types of triangles, whether it is scalene, isosceles or equilateral. To be noted, the base and height of the triangle are perpendicular to each other. The unit of area is measured in square units ($m^2, {cm}^2$).

Formula used:
Area of a triangle = \[\dfrac{1}{2}|\mathop A \limits^ \to \times \mathop B\limits^ \to |\]
A and B are the vectors.

Note:
The area of a triangle is defined as the total region that is enclosed by the three sides of any particular triangle. Basically, it is equal to half of the base times height, i.e., \[A = \dfrac{1}{2} \times base \times height\]. Hence, to find the area of a tri-sided polygon, we have to know the base (b) and height (h) of it.