Answer :
Answer:
Distance from point (0,1,1) to the given line is zero.
Step-by-step explanation:
Given parametric equations of line,
x=2t, y=5-2t, z=1+t
To find distance from (0,1,1), we have to eliminate t from above equations so that,
[tex]y=5-2t=5-x\implies x+y=5=4+1=4+z-t=4+z-\frac{1}{2}x[/tex]
[tex]\implies 3x+2y-2z-8=0\hfill (1)[/tex]
whose direction ratioes are (l,m,n)=(3,2,-2) and distance fro point (a,b,c)=(0,1,1) is given by,
[tex]\frac{al+bm+cn}{\sqrt{l62+m^2+n^2}}=\frac{(3\times 0)+(2\times 1)+(-2)(1)}{\sqrt{3^2+2^2+(-2)^2}}=\frac{0+2-2}{\sqrt{17}}=0[/tex]
Distance between point (0,1,1) and (1) is zero. That is point 90,1,1) is lies on the line (1).
The distance between points, is how far the points are, from one another,
The distance between point P and line L is 0 units
The given parameters are:
[tex]x = 2t[/tex]
[tex]y = 5 - 2t[/tex]
[tex]z = 1 +t[/tex]
Make t the subject in [tex]x = 2t[/tex]
[tex]t = \frac 12 x[/tex]
Substitute [tex]t = \frac 12 x[/tex] in [tex]y = 5 - 2t[/tex] and [tex]z = 1 +t[/tex]
[tex]y = 5 - 2 \times \frac 12x[/tex]
[tex]y = 5 - x[/tex]
[tex]x + y = 5[/tex]
[tex]z = 1 + \frac 12x[/tex]
[tex]z - \frac 12x = 1[/tex]
Express [tex]5[/tex] as [tex]4 + 1[/tex] in [tex]x + y = 5[/tex]
[tex]x + y = 4 + 1[/tex]
[tex]x + y - 4 = 1[/tex]
Substitute [tex]z - \frac 12x[/tex] for 1 in [tex]x + y - 4 = 1[/tex]
[tex]x + y - 4 = z - \frac 12x[/tex]
Multiply through by 2
[tex]2x + 2y - 8 = 2z - x[/tex]
Collect like terms
[tex]x + 2x + 2y -2z - 8 = 0[/tex]
[tex]3x + 2y -2z - 8 = 0[/tex]
So, the distance ratio
[tex](l,m,n) = (x,y,z)[/tex]
Replace x, y and z with their coefficients
[tex](l,m,n) = (3,2,-2)[/tex]
From the question, we have:
[tex](a,b,c)=(0,1,1)[/tex]
So, the distance (d) is:
[tex]d = \frac{|a \times b|}{|a|}[/tex]
This gives
[tex]d = \frac{al + bm + cn}{\sqrt{l^2 + m^2 + n^2}}[/tex]
Substitute known values
[tex]d = \frac{0 \times 3 + 1 \times 2 + 1 \times -2}{\sqrt{3^2 + 2^2 + (-2)^2}}[/tex]
[tex]d = \frac{0}{\sqrt{17}}[/tex]
[tex]d = 0[/tex]
Hence, the distance between point P and line L is 0 units
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