Question

OABCDE is a regular hexagon of side 2 units in the XY plane in the first quadrant. O being the origin and OA taken along the x axis. A point P is taken on a line parallel to the z axis through the centre of the hexagon at a distance of 3 units from O in the positive Z direction. Then find the vector $\overrightarrow{AP}$.

Answer 1

Hint: Now we have a regular hexagon of side 2 units. We will first take the midpoint of OA and of length 1 unit. Now let X be the center of hexagon and M be the midpoint OA. Since we have coordinates of O and A and we know that the triangle OXM is a right angle triangle we will find the length of XM by taking trigonometric ratio. Now since we have distance OM we can write the coordinates of point X. Hence we can also write the coordinates of point P as the x and y coordinate will be the same. Now we know the distance of origin from the point P. hence using this we will find the z coordinate of point P. Now we know the coordinates of P and A hence we can write the vector AP as $\overrightarrow{AP}=\overrightarrow{P}-\overrightarrow{A}$ .

Complete step by step solution:
Now we have that AOBCDE as a regular hexagon with side 2 units in XY plane.
Now O is origin hence O is nothing but $0\hat{i}+0\hat{j}+0\hat{k}$
Since OA is along x axis we can say that the equation form A is $2\hat{i}$
Let X be the centre of hexagon and M be the mid point of OA. Hence the coordinates of A will be (1, 0, 0) since each side is 2 units.
Now let us first draw the figure with the problem.

Now since the hexagon is a regular hexagon we can say that $\angle OXA={{60}^{\circ }}$
Hence we have that \[\angle OXM={{30}^{\circ }}\]
Now we know that OM = 1 unit.
And by taking
$\begin{align}
  & \Rightarrow \tan \left( \angle OXM \right)=\dfrac{OM}{XM} \\
 & \Rightarrow \tan \left( 30 \right)=\dfrac{1}{XM} \\
\end{align}$
$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{1}{XM}$
Hence we get XM = $\sqrt{3}$ .
Now we know that point X is in XY plane and hence its Z coordinate is 0.
Hence we get the coordinates of X as $\left( 1,\sqrt{3},0 \right)$ .
Now we know that point P is along the line passing through X and parallel to Z axis.
Let z coordinate of P be $\lambda $
Hence we get the coordinate of P as $\left( 1,\sqrt{3},\lambda \right)$
Now we know that the distance of P from origin is 3 units.
Hence we get
$\begin{align}
  & \Rightarrow {{1}^{2}}+{{\left( \sqrt{3} \right)}^{2}}+{{\lambda }^{2}}={{3}^{2}} \\
 & \Rightarrow 1+3+{{\lambda }^{2}}=9 \\
 & \Rightarrow {{\lambda }^{2}}=5 \\
\end{align}$
Hence the value of $\lambda =\sqrt{5}$ . Since $\lambda $ is along positive z axis.
Hence we get coordinate of P is $\left( 1,\sqrt{3},\sqrt{5} \right)$
 Now coordinate of A is $\left( 2,0,0 \right)$ Hence the equation of vector AP is
$\begin{align}
  & \Rightarrow \left( 1-2 \right)\hat{i}+\left( \sqrt{3}-0 \right)\hat{j}+\left( \sqrt{5}-0 \right)\hat{k} \\
 & \Rightarrow -\hat{i}+\sqrt{3}\hat{j}+\sqrt{5}\hat{k} \\
\end{align}$
Hence the equation of AP vector is $-\hat{i}+\sqrt{3}\hat{j}+\sqrt{5}\hat{k}$ .

Note: Now note that the regular hexagon is a symmetric structure. Now if we join X to all the edges of Hexagon we will get 6 angles. Since Hexagon is a symmetric structure all these 6 angles will be equal. Also these 6 angles complete one rotation and hence their sum is ${{360}^{\circ }}$ . Hence we have $6x={{360}^{\circ }}\Rightarrow x={{60}^{\circ }}$ . Hence we have the $\angle OXA$as ${{60}^{\circ }}$ .