Question

If $\overrightarrow a ,\overrightarrow b $ vectors perpendicular to each other and $\left| {\overrightarrow a } \right| = 2,\left| {\overrightarrow b } \right| = 3,\overrightarrow c \times \overrightarrow a = \overrightarrow b $then the least value of $2\left| {\overrightarrow c - \overrightarrow a } \right|$ is

Answer 1

Hint: To find the value of 2$\left| {\overrightarrow c - \overrightarrow a } \right|$, calculate the value of ${\left| {\overrightarrow c - \overrightarrow a } \right|^2}$ and take square root of that value and multiply it with 2.
You can find the value of ${\left| {\overrightarrow c - \overrightarrow a } \right|^2}$by the formula given below:
$ \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = \left( {\overrightarrow c - \overrightarrow a } \right)\left( {\overrightarrow c - \overrightarrow a } \right)$

Complete step by step answer:

Let us see what is given in the question, we have given the value of $|\overrightarrow a |\& |\overrightarrow b |$as follows
$ \Rightarrow |\overrightarrow a | = 2$and
$ \Rightarrow |\overrightarrow b | = 3$
First of all, find the value of ${\left| {\overrightarrow c - \overrightarrow a } \right|^2}$i.e.
$ \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = \left( {\overrightarrow c - \overrightarrow a } \right)\left( {\overrightarrow c - \overrightarrow a } \right)$
By opening the bracket and applying distributive property we get,
$
   \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = \overrightarrow c \left( {\overrightarrow c - \overrightarrow a } \right) - \overrightarrow a \left( {\overrightarrow c - \overrightarrow a } \right) \\
   \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = \overrightarrow c .\overrightarrow c - \overrightarrow c .\overrightarrow a - \overrightarrow a .\overrightarrow c + \overrightarrow a .\overrightarrow a \\
$
As $\overrightarrow a .\overrightarrow c = \overrightarrow c .\overrightarrow a $then we get,
$ \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = \overrightarrow c .\overrightarrow c - \overrightarrow c ..\overrightarrow a - \overrightarrow c .\overrightarrow a + \overrightarrow a \overrightarrow a $
$ \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = {\overrightarrow {\left| c \right|} ^2} - 2\overrightarrow c .\overrightarrow a + {\overrightarrow {\left| a \right|} ^2}$…….(1)
Now, we have to find the value of $\overrightarrow a .\overrightarrow c $as given below
The dot product of a and c is given by
$ \Rightarrow \overrightarrow c .\overrightarrow a = |\overrightarrow c ||\overrightarrow a |\cos \theta $, where $\theta $ is the angle between a and c.
Putting the value $|\overrightarrow a | = 2$ in this we get,
$ \Rightarrow \overrightarrow c .\overrightarrow a = 2|\overrightarrow c |\cos \theta $
Now find the value of ${\left| {\overrightarrow c } \right|^{}}$by scalar product i.e.
$ \Rightarrow \left| {\overrightarrow b } \right| = \left| {\overrightarrow c } \right| \times \left| {\overrightarrow a } \right| = \left| {\overrightarrow c } \right|\left| {\overrightarrow a } \right|\sin \theta $
Putting the value of $|\overrightarrow a |\& |\overrightarrow b |$ we get,
$ \Rightarrow 3 = 2.\left| {\overrightarrow c } \right|\sin \theta $
$ \Rightarrow \left| {\overrightarrow c } \right| = \dfrac{3}{{2\sin \theta }}$
Put the value of $|\overrightarrow a |\& |\overrightarrow c |$ in equation (1) we get,
$ \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = {\overrightarrow {\left| c \right|} ^2} - 2\overrightarrow c .\overrightarrow a + {\overrightarrow {\left| a \right|} ^2} = {\overrightarrow {\left| c \right|} ^2} - 2.\overrightarrow {\left| c \right|} \overrightarrow {\left| a \right|} \cos \theta + {\overrightarrow {\left| a \right|} ^2}$=
$ \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = {(\dfrac{3}{{2\sin \theta }})^2} - 2(\dfrac{3}{{2\sin \theta }}).2\cos \theta + {2^2}$
$
   \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = (\dfrac{9}{{4{{\sin }^2}\theta }}) - (\dfrac{3}{{\sin \theta }}).2\cos \theta + {2^2} \\
   \Rightarrow {\left| {\overrightarrow c - \overrightarrow a } \right|^2} = \dfrac{9}{{4{{\sin }^2}\theta }} - \dfrac{{6\cos \theta }}{{sin\theta }} + 4 \\
   \Rightarrow \left| {\overrightarrow c - \overrightarrow a } \right| = \sqrt {\dfrac{9}{{4{{\sin }^2}\theta }} - \dfrac{{6\cos \theta }}{{sin\theta }} + 4} \\
   \Rightarrow 2\left| {\overrightarrow c - \overrightarrow a } \right| = 2\sqrt {\dfrac{9}{{4{{\sin }^2}\theta }} - \dfrac{{6\cos \theta }}{{sin\theta }} + 4} \\
 $
$
\Rightarrow 2\left| {\overrightarrow c - \overrightarrow a } \right| = \sqrt {\dfrac{{4 \times 9}}{{4{{\sin }^2}\theta }} - \dfrac{{4 \times 6 \times \cos \theta }}{{sin\theta }} + 4 \times 4} \\
   \Rightarrow 2\left| {\overrightarrow c - \overrightarrow a } \right| = \sqrt {\dfrac{9}{{{{\sin }^2}\theta }} - \dfrac{{24 \times \cos \theta }}{{sin\theta }} + 16} = \sqrt {9\cos e{c^2}\theta - 24 \times \cot \theta + 16} \\
 $
As $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta $, $\dfrac{1}{{{{\sin }^2}\theta }} = \cos e{c^2}\theta $ and $\cos e{c^2}\theta = {\cot ^2}\theta + 1$ we get,
$
   \Rightarrow 2\left| {\overrightarrow c - \overrightarrow a } \right| = \sqrt {9(1 + {{\cot }^2}\theta ) - 24 \times \cot \theta + 16} = \sqrt {9 + 16 + 9{{\cot }^2}\theta - 24\cot \theta } \\
   \Rightarrow 2\left| {\overrightarrow c - \overrightarrow a } \right| = \sqrt {9{{\cot }^2}\theta - 24\cot \theta + 25} \\
 $
As, the minimum value of $\cot \theta $ is zero at $\theta = \dfrac{\pi }{2}$. Therefore,
$ \Rightarrow 2\left| {\overrightarrow c - \overrightarrow a } \right| = \sqrt {9 \times {{\cot }^2}\dfrac{\pi }{2} - 24\cot \dfrac{\pi }{2} + 25} $
As the minimum value of this vector can be obtained by putting $\theta = \dfrac{\pi }{2}$.
$ \Rightarrow 2\left| {\overrightarrow c - \overrightarrow a } \right| = \sqrt {25} = 5$
Hence, 5 is the least value of $2\left| {\overrightarrow c - \overrightarrow a } \right|$.

Note: Some students get confused in cross and dot product as in the question it is already given that cross product of a and c is b. Some students consider it as a dot product and take cosine function. Your answer can get wrong. Secondly, while taking square we will take two vectors and we consider it as a dot product and use the cosine function. Take care of these things.
Types of vectors are given as follows:
Zero or Null Vector: When starting and ending points of a vector are the same is called zero or null vector.
Unit Vector: If the magnitude of a vector is unity then the vector is called a unit vector.
Free Vectors: When the initial point of a vector is not defined then those types of vectors are said to be free vectors.
Negative of a Vector: A vector is said to be a negative vector if the magnitude of a vector is the same as the given vector but the direction is opposite to it.
Like and Unlike Vectors: Unlike vectors are the vectors whose direction is opposite to each other but the direction of both is the same in like vectors.