Question

The extremities of a diagonal of a rectangular parallelopiped whose faces are parallel to the reference planes are (- 2,4,6) and (3,16,13). The length of the base diagonal is
1)13. 2)√13. 3)2√13. 4)169​

Answer 1

Answer:

$\boxed{\bf \: Length\:of\:base \: diagonal = 13 \: units \: }$

Step-by-step explanation:

Given that, extremities of a diagonal of a rectangular parallelopiped whose faces are parallel to the reference planes are (- 2, 4, 6) and (3, 16, 13).

As we have to find the diagonal of the base.

So, The coordinates of the base diagonal in the XY plane be (- 2, 4, 6) and (3, 16, 6)

We know,

Distance Formula

Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) be two points in the cartesian plane, then distance between A and B is given by

$\boxed{\tt{ AB= \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1}) }^{2} }}}$

So, using distance formula, the length of diagonal is

$\sf \: Length\:of\:base \: diagonal$

$\sf \: = \: \sqrt{ {[ 3 - ( - 2)]}^{2} + {(16 - 4)}^{2} + {(6 - 6)}^{2} }$

$\sf \: = \: \sqrt{ { {(3 + 2)}^{2} } + {(12)}^{2} + {(0)}^{2} }$

$\sf \: = \: \sqrt{ { {(5)}^{2} } + 144 + 0 }$

$\sf \: = \: \sqrt{ 25+ 144}$

$\sf \: = \: \sqrt{169}$

$\sf \: = \: \sqrt{13 \times 13}$

$\sf \: = \: 13 \:$

Hence,

$\implies\sf \:\boxed{\bf \: Length\:of\:base \: diagonal = 13 \: units \: }$