Question

How do you simplify \[\dfrac{4}{5} - \dfrac{1}{2}\]

Answer 1

Hint: Here in this question, we have - symbol which represents the subtraction and we have to subtract the two numbers. The numbers are in the form of fraction. by taking the LCM for the denominators and we are going to simplify the given numbers.

Complete step-by-step solution:
A vector that has a magnitude of 1 is a unit vector. It is also known as Direction Vector.
 The given points are A(3, -1, 2), B(1, -1, -3) and C(4, -3, 1) lie on the plane ABC. Therefore \[AB\] and \[AC\]are the vectors which is on the plane ABC. Then \[\overrightarrow {AB} \times \overrightarrow {AC} \]is perpendicular to the plane.
Then the unit vector is determined by using the formula \[\dfrac{{\overrightarrow {AB} \times \overrightarrow {AC} }}{{||\overrightarrow {AB} \times \overrightarrow {AC} ||}}\]----- (1)
The vector \[\overrightarrow {AB} \] is determined by the \[B - A\], substituting the values of A and B we get
\[ \Rightarrow \overrightarrow {AB} = (1, - 1, - 3) - (3, - 1,2)\]
\[
   \Rightarrow \overrightarrow {AB} = (1 - 3, - 1 - ( - 1), - 3 - 2) \\
   \Rightarrow \overrightarrow {AB} = (1 - 3, - 1 + 1, - 3 - 2) \\
 \]
\[ \Rightarrow \overrightarrow {AB} = ( - 2,0, - 5)\] ---------- (2)
The vector \[\overrightarrow {AC} \] is determined by the \[C - A\], substituting the values of A and C we get
\[ \Rightarrow \overrightarrow {AB} = (4, - 3,1) - (3, - 1,2)\]
\[
   \Rightarrow \overrightarrow {AC} = (4 - 3, - 3 - ( - 1),1 - 2) \\
   \Rightarrow \overrightarrow {AB} = (4 - 3, - 3 + 1,1 - 2) \\
 \]
\[ \Rightarrow \overrightarrow {AB} = (1, - 2, - 1)\] ------------ (3)
The \[\overrightarrow {AB} \times \overrightarrow {AC} \] is a cross product. So we have
\[ \Rightarrow \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  { - 2}&0&{ - 5} \\
  1&{ - 2}&{ - 1}
\end{array}} \right|\]
On simplifying
\[
   \Rightarrow \overrightarrow {AB} \times \overrightarrow {AC} = i(0( - 1) - ( - 2)( - 5)) - j(( - 2)( - 1) - (1)( - 5)) \\
  k(( - 2)( - 2) - (1)(0)) \\
 \]
\[ \Rightarrow \overrightarrow {AB} \times \overrightarrow {AC} = i(0 + 10) - j(2 + 5) + k(4 - 0)\]
\[ \Rightarrow \overrightarrow {AB} \times \overrightarrow {AC} = 10i - 7j + 4k = (10, - 7,4)\]---- (4)
The \[\left\| {\overrightarrow {AB} \times \overrightarrow {AC} } \right\|\] is determined by
\[
   \Rightarrow \left\| {\overrightarrow {AB} \times \overrightarrow {AC} } \right\| = \sqrt {{{10}^2} + {{( - 7)}^2} + {4^2}} \\
   \Rightarrow \left\| {\overrightarrow {AB} \times \overrightarrow {AC} } \right\| = \sqrt {100 + 49 + 16} \\
   \Rightarrow \left\| {\overrightarrow {AB} \times \overrightarrow {AC} } \right\| = \sqrt {165} \\
 \]
Therefore the unit vector is given by
\[\dfrac{{\overrightarrow {AB} \times \overrightarrow {AC} }}{{||\overrightarrow {AB} \times \overrightarrow {AC} ||}} = \left( {\dfrac{{10}}{{\sqrt {165} }},\dfrac{{ - 7}}{{\sqrt {165} }},\dfrac{4}{{\sqrt {165} }}} \right)\]
Hence this is the unit vector perpendicular to the plane

Note: The vectors are multiplied by the two kinds one is dot product and the other one is cross product. The dot product is like multiplication itself. The terms are multiplied which are in the same coordinate. But in case of cross product while multiplying the terms we consider the determinant for the points or vector.