Question

A parallelepiped is formed by planes drawn through the points $P(6,8,10)$ and $Q(3,4,8)$ parallel to the coordinate planes. Find the length of edges and diagonals of the parallelepiped.

Answer 1

Hint: For solving this question first we will draw planes through the points $P(6,8,10)$ and $Q(3,4,8)$ parallel to the coordinate planes. After that, we will prove that parallelepiped formed by the given planes will be a cuboid and we will find its dimensions. Moreover, we will use the formula $\sqrt{l^2+b^2+h^2}$ to find the body diagonal of the cuboid of length $l$ , breadth $b$ and height $h$ .

Complete step by step solution:
Given:
We have to find the length of edges and diagonals of the parallelepiped formed by planes drawn through the points $P(6,8,10)$ and $Q(3,4,8)$ parallel to the coordinate planes.
Now, before we proceed we should know that if there is a point $A(x_1,y_2,z_2)$ then $x=x_1$ , $y=y_1$ and $z=z_1$ planes will pass through point $A(x_1,y_2,z_2)$ and parallel to the coordinate planes.
Now, from the above discussion, we conclude that, $x=6$ , $y=8$ and $z=10$ planes will be passing through point $P(6,8,10)$ and parallel to the coordinate planes. For more clarity look at the figure given below:

Now, from the above discussion, we conclude that, $x=3$ , $y=4$ and $z=8$ planes will be passing through point $Q(3,4,8)$ and parallel to the coordinate planes. For more clarity look at the figure given below:

Now, there are the following six planes:
$\begin{array}{rl}& x=6\text{ and }x=3\\ & y=8\text{ and y}=4\\ & z=10\text{ and }z=8\end{array}$
Now, parallelepiped formed by the above six planes will be a cuboid. For more clarity look at the figure given below:

Now, to find the dimensions of the cuboid we will find the distance between the following planes:
$\begin{array}{rl}& x=6\text{ and }x=3\\ & y=8\text{ and y}=4\\ & z=10\text{ and }z=8\end{array}$
Now, as the above three pairs of planes are parallel so, we will simply find the difference in their values.
For length find the distance between planes $y=8$ and $y=4$ , for breadth find the distance between planes $x=6$ and $x=3$ , and for height find the distance between planes $z=10$ and $z=8$ Then,
$\begin{array}{rl}& Length=l=8-4=4\\ & Breadth=b=6-3=3\\ & Height=h=10-8=2\end{array}$
Now, from the above result, we conclude that the length, breadth and height of the cuboid will be 4, 3 and 2 units respectively.
Now, as we know that length of the body diagonal of the cuboid of length $l$ , breadth $b$ and height $h$ is $\sqrt{l^2+b^2+h^2}$ . Then,
Length of the body diagonal $=\sqrt{l^2+b^2+h^2}=\sqrt{4^2+3^2+2^2}=\sqrt{16+9+4}=\sqrt{29}$ units.
Now, we conclude that the required parallelepiped will be a cuboid of length, breadth and height of the cuboid will be 4, 3 and 2 units respectively and length of the diagonal of the parallelepiped will be $\sqrt{29}\approx 5.385$ units.

Note: The coordinates of the vertices of the parallelepiped can be found out by finding the point of intersection of the planes taken 3 at a time. For example, the point of intersection of the planes x=6, y=8 and z=10 is (6,8,10), which is the coordinate of the vertex P.