Question

the distance of the line x+3=y+4=z+5 from the origin is​

Answer 1

SOLUTION

TO DETERMINE

The distance of the line x + 3 = y + 4 = z + 5 from the origin

CONCEPT TO BE IMPLEMENTED

If equation of any line is

$\displaystyle\sf{ \frac{x - a}{l} = \frac{y - b}{m} = \frac{z - c}{n} }$

$\sf{Then \: the \: distance \: of \: the \: line \: from \: the \: point \: P( \alpha , \beta , \gamma ) \: \: is }$

$\displaystyle\sf{ = \sqrt{ \bigg[ \: {( \alpha - a)}^{2} + {( \beta - b)}^{2} + {( \gamma - c)}^{2} - \frac{ \{ \: l( \alpha - a) + m( \beta - b) + n( \gamma - c)\}}{ {l}^{2} + {m}^{2} + {n}^{2} } \: \bigg]} }$

EVALUATION

Here the given equation of the line is

x + 3 = y + 4 = z + 5

Comparing with the general equation

$\displaystyle\sf{ \frac{x - a}{l} = \frac{y - b}{m} = \frac{z - c}{n} }$

We get

a = - 3 , b = - 4 , c = - 5

l = m = n = 1

Now the given point is origin

$\sf{ \alpha = 0, \beta = 0, \gamma = 0}$

Hence the required distance

$\displaystyle\sf{ = \sqrt{ \bigg[ \: {( 0 + 3)}^{2} + {( 0 + 4)}^{2} + {( 0 + 5)}^{2} - \frac{ \{ \: 1.( 0 + 3) + 1.( 0 + 4) + 1.( 0 + 5)\}}{ {1}^{2} + {1}^{2} + {1}^{2} } \: \bigg]} \: \: \: unit }$

$\displaystyle\sf{ = \sqrt{ \bigg[ \: {3}^{2} + {4}^{2} + {5}^{2} - \frac{ ( \: 3 + 4 + 5)}{ 3 } \: \bigg]} \: \: \: unit }$

$\displaystyle\sf{ = \sqrt{ \bigg[ 50 - \frac{ 12}{ 3 } \: \bigg]} \: \: \: unit }$

$\displaystyle\sf{ = \sqrt{ \bigg[ 50 - 4 \: \bigg]} \: \: \: unit }$

$\displaystyle\sf{ = \sqrt{ 46} \: \: \: unit }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. a butterfly is moving in a straight line in the space.

let this path be denoted by a line l whose equation is x-2/2=2-y...

https://brainly.in/question/30868719

2. Find the distance of point p(3 4 5) from the yz plane

https://brainly.in/question/8355915