Question

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

Answer 1

Concept

Distance of a line from the origin in 3d plane can be calculated by finding the derivative of the distance equation.

Given

Equation of line x+3=y+4=z+5 and origin (0,0,0).

Find

Distance between line and origin.

Solution

The equation of line is given as x+3=y+4=z+5.

consider x+3=y+4=z+5 = k .

now the coordinates changes to P( t-3, t-4 , t-5 )

Distance between new point and origin ,

OP=$\sqrt{(t-3)^{2} +(t-4)^{2}+(t-5)^{2} }$

now smallest distance can be calculated using maxima and minima,

$\frac{d(\sqrt[]{(t-3)^{2} +(t-4)^{2} +(t-5)^{2} }) }{dt}$ =0

$d((t-3)^{2}+(t-4) ^{2}+(t-5)^{2})$=0

2(t-3)+2(t-4)+2(t-5)=0

9t-6-8-10=0

9t=-24

t=-24/9

t=-8/3

substituting in above distance equation

OP =$\sqrt{(t-3)^{2} +(t-4)^{2}+(t-5)^{2} }$

OP=11.633

The distance of the line x+3=y+4=z+5 from the origin is​ 11.633.

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