Question

If the vectors \[\left( {\hat i + \hat j + \hat k} \right)\] and \[3\hat i\]forms two sides of a triangle, then area of the triangle is
(1) \[\sqrt 3 \]\[unit\]
(2) \[2\sqrt 3 \]\[unit\]
(3) \[\dfrac{3}{{\sqrt 2 }}\]\[unit\]
(4) \[3\sqrt 2 \]\[unit\]

Answer 1

Hint: If two sides of a triangle are given in vector form, then the area of the triangle is given by half time of the magnitude of the cross product of two sides. If \[\overrightarrow a \] and \[\overrightarrow b \] are the vectors representing the side of a triangle then its area is given by\[\dfrac{1}{2}\left| {\overrightarrow a \times \overrightarrow b } \right|\].
Cross product: Multiplication of vectors can be given in two ways. First way is a scalar or dot product. Second one is known as a vector product or cross product.
Cross product of two vectors gives the area vector of a parallelogram formed by the two vectors. Then the area of the triangle is half of the area vector of a parallelogram.

Complete step by step answer:
Given: two sides of a triangle. Let PQR be the triangle. Then two sides of triangle is given by
\[\overrightarrow {PQ} = \left( {\hat i + \hat j + \hat k} \right)\] and \[\overrightarrow {QR} = 3\hat i\]
Area of triangle in vector form is given by
= \[\dfrac{1}{2}\]\[\left( {magnitude{\text{ }}cross{\text{ }}product{\text{ }}of{\text{ }}two{\text{ }}vectors} \right)\]
=\[\dfrac{1}{2}\left| {\overrightarrow {PQ} \times \overrightarrow {QR} } \right|\]
=\[\dfrac{1}{2}\left| {\left( {\hat i + \hat j + \hat k} \right) \times 3\hat i} \right|\]
=\[\dfrac{1}{2}\left| {\left( {\hat i + \hat j + \hat k} \right) \times \left( {3\hat i + 0\hat j + 0\hat k} \right)} \right|\]
\[\overrightarrow {PQ} \times \overrightarrow {QR} \]=\[\left( {\hat i + \hat j + \hat k} \right) \times 3\hat i\]
= \[\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  1&1&1 \\
  3&0&0
\end{array}} \right|\]
=\[\hat i(0 - 0) - \hat j(0 - 3) + \hat k(0 - 3)\]
=\[3\hat j - 3\hat k\]
Hence, area of triangle PQR= \[\dfrac{1}{2}\left| {3\hat j - 3\hat k} \right|\]
=\[\dfrac{1}{2}\left( {\sqrt {{3^2} + {{( - 3)}^2}} } \right)\]
=\[\dfrac{{3\sqrt 2 }}{2}\]
=\[\dfrac{3}{{\sqrt 2 }}\]\[unit\]

So, the correct answer is “Option C”.

Note:
For cross product, vectors should be written in the form of\[\left( {x\hat i + y\hat j + z\hat k} \right)\]. if any two coordinates are given then consider the third coordinate as zero. Because for a cross product of two vectors all three coordinates \[\left( {x,y,z} \right)\]are necessary. Otherwise we don’t get the correct answer. Students make mistakes in writing a vector in the form of \[\left( {x\hat i + y\hat j + z\hat k} \right)\]. In question the second vector has only one coordinate. so we should write this vector in the form of\[\left( {x\hat i + y\hat j + z\hat k} \right)\]. Remember one thing that cross product of two vectors can only defined in three dimensional coordinate plane