Question

A mosquito flies from the hole in a mosquito net top corner diametrically opposite. If the net is 3 × 2 × 2 m then the displacement of the mosquito is:
A) $\sqrt {13} m$
B) $\sqrt {17} m$
C) $\sqrt {11} m$
D) None of these

Answer 1

Hint: We will use the concept of vectors for this question. The dimensions of the net are given, these will be the displacement vectors of net magnitude of which will give the displacement of the mosquito from hole.
Displacement vector is used to show the change in position of the vector with respect to time

Complete step by step answer:
The given dimensions of net are:
3 × 2 × 2 m

The displacement vector can be written in the form:
$\vec d$ = $x\hat i + y\hat j + z\hat k$
The displacement of the mosquito is given by OD since it flies diametrically opposite (given)
Substituting the values:
$\vec d$ = \[\vec O\vec D = \left( {0\hat i + 0\hat j + 0\hat k} \right) + \left( {3\hat i + 2\hat j + 2\hat k} \right)\]
$\vec d$ = $3\hat i + 2\hat j + 2\hat k$
Since O is the origin with points (0 , 0 , 0) and AD diagonal has all the 3 dimensions (x , y and z) equal to the dimensions of the net.
The displacement of the mosquito is given by this position vector’s magnitude:
$\left| {x\hat i + y\hat j + z\hat k} \right| = \sqrt {{x^2} + {y^2} + {z^2}} $
\[\left| {3\hat i + 2\hat j + 2\hat k} \right| = \sqrt {{{(3)}^2} + {{(2)}^2} + {{(2)}^2}} \]
\[\left| {3\hat i + 2\hat j + 2\hat k} \right|\] = $\sqrt {9 + 4 + 4} $
   \[\left| {3\hat i + 2\hat j + 2\hat k} \right|\] = $\sqrt {17} $
$\vec d$ = $\sqrt {17} $


As the dimensions are in meters (m), this displacement will also be in m
$\sqrt {17} m$
Therefore, the displacement of the mosquito that flies from the hole in a mosquito net is $\sqrt {17} m$

Note:Vectors have both magnitude and direction.
$\hat i,\hat j,\hat k$ are unit vectors of respective unit vectors of x , y and z axis.
The different representations are :
$\vec a$ = shows that ‘a’ is a vector
$\hat a$ = shows that ‘a’ is a unit vector
$\left| a \right|$ = denotes the magnitude of ‘a’