Question

Obtain coordinates of points on the line x/2=y/2=z/1, which are at 6 unit distance. ​

Answer 1

Answer:

x+1

=

3

y+2

=

6

z−3

Let the point be

=(2k−1,3k−2,6k+3)

Distance from (1,−2,3)

=

(2k−1−1)

2

+(3k−2+2)

2

+(6k+3−3)

2

=

(2k−2

2

+9k

2

+36k

2

=3

∴4k

2

+4−8k+45k

2

=9

∴49k

2

−8k−5=0

∴k=

98

8±

64+20×49

∴k=

98

8±

65+980

∴k=

98

8±32

=

98

40

=

49

20

∴ Point are (

49

−9

,

49

−38

,

49

267

)

I HOPE MY ANSWER HELPED YOU PLEASE MARK AS BRAINLIEST

Answer 2

The points on the line that are at a distance of 6 units from the origin are:

(x,y,z) = 2(2,2,1) = (4,4,2) and (-4,-4,-2)

The line's equation is provided by:

x/2 = y/2 = z/1

Let's choose a point on the line as the origin, which is (0,0,0) in Cartesian coordinates. Then any point on the line can be represented as a scalar multiple of the direction vector of the line, which is (2,2,1). That is, any point on the line can be written as:

(x,y,z) = t(2,2,1)

where t is a scalar parameter.

Now we need to find the points on this line that are at a distance of 6 units from the origin. The formula for the distance between the two positions (x1,y1,z1) and (x2,y2,z2) is:

distance = $sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)$

So we need to solve the following equation for t:

$sqrt((2t)^2 + (2t)^2 + (t)^2) = 6$

Simplifying, we get:

$sqrt(9t^2) = 6\\3t = 6\\t = 2$

Therefore, the points on the line that are at a distance of 6 units from the origin are:

(x,y,z) = 2(2,2,1) = (4,4,2) and (-4,-4,-2)

These are the coordinates of the two points on the line that are at a distance of 6 units from the origin.

for such more question on coordinate

https://brainly.in/question/23880087

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