Question

If \[\left| {\vec a} \right| = 2;\left| {\vec b} \right| = 3\,and\,\vec a.\vec b = 4,\,find\,\left| {\vec a - \vec b} \right|\] ?

Answer 1

Hint: Here the given question is of vector algebra, and magnitude of two vectors, are given, as well as their dot product magnitude is also given, here we are asked to find the magnitude of their difference, so first we need to assume the desired vectors and then solve accordingly.

Complete step by step answer:
Here in the given question we are provided with the dot product of vectors, and the magnitude of the two vectors, let suppose the two vectors as:
\[ \Rightarrow \vec a = {x_1}\hat i + {x_2}\hat j + {x_3}\hat k\]and
\[ \Rightarrow \vec b = {y_1}\hat i + {y_2}\hat j + {y_3}\hat k\]
Now we need to find:
\[ \Rightarrow \left| {\vec a - \vec b} \right|\]
Expressing this in assumed vector we get:
\[
   \Rightarrow \left| {\vec a - \vec b} \right| = \left| {\left( {{x_1}\hat i + {x_2}\hat j + {x_3}\hat k} \right) - \left( {{y_1}\hat i + {y_2}\hat j + {y_3}\hat k} \right)} \right| \\
   \Rightarrow \left| {\vec a - \vec b} \right| = \left| {\left( {({x_1} - {y_1})\hat i + ({x_2} - {y_2})\hat j + ({x_3} - {y_3})\hat k} \right)} \right| \\
 \]
Writing the magnitude we get:
\[
   \Rightarrow \left| {\vec a - \vec b} \right| = \sqrt {{{({x_1} - {y_1})}^2} + {{({x_2} - {y_2})}^2} + {{({x_3} - {y_3})}^2}} \\
   \Rightarrow \left| {\vec a - \vec b} \right| = \sqrt {{x_1}^2 + {y_1}^2 - 2{x_1}{y_1} + {x_2}^2 + {y_2}^2 - 2{x_2}{y_2} + {x_3}^2 + {y_3}^2 - 2{x_3}{y_3}} \\
   \Rightarrow \left| {\vec a - \vec b} \right| = \sqrt {\left( {{x_1}^2 + {x_2}^2 + {x_3}^2} \right) + \left( {{y_1}^2 + {y_2}^2 + {y_3}^2} \right) - 2\left( {{x_1}{y_1} + {x_2}{y_2} + {x_3}{y_3}} \right)} \\
 \]
Now we know from the given magnitude of vectors we have:
\[
   \Rightarrow \left| {\vec a} \right| = 2 = \sqrt {\left( {{x_1}^2 + {x_2}^2 + {x_3}^2} \right)} \\
   \Rightarrow \left( {{x_1}^2 + {x_2}^2 + {x_3}^2} \right) = {2^2} = 4 \\
 \]
And
\[
   \Rightarrow \left| {\vec b} \right| = 3 = \sqrt {\left( {{y_1}^2 + {y_2}^2 + {y_3}^2} \right)} \\
   \Rightarrow \left( {{y_1}^2 + {y_2}^2 + {y_3}^2} \right) = {3^2} = 9 \\
 \]
Now from the magnitude of the dot product we have:
\[ \Rightarrow \vec a.\vec b = 4 = \left( {{x_1}{y_1} + {x_2}{y_2} + {x_3}{y_3}} \right)\]
Putting these obtained values in the main equation of required magnitude we have:
\[
   \Rightarrow \left| {\vec a - \vec b} \right| = \sqrt {\left( {{x_1}^2 + {x_2}^2 + {x_3}^2} \right) + \left( {{y_1}^2 + {y_2}^2 + {y_3}^2} \right) - 2\left( {{x_1}{y_1} + {x_2}{y_2} + {x_3}{y_3}} \right)} \\
   \Rightarrow \left| {\vec a - \vec b} \right| = \sqrt {4 + 9 - 2 \times 4} = \sqrt {13 - 8} = \sqrt 5 \\
 \]
Here we get the required magnitude of the vector as asked in the question.

Note: Here in the given question, we need to find a new vector magnitude, and for this kind of question we first need to find the required vector, then on applying the formulae of magnitude we can get the required magnitude, and the given data will be used accordingly.