Answer:

To find the coordinates of points on the line with the equation x/1 = y-1/2 = z-1/2 = 1/2 at a distance of √11 units from the origin, we need to solve for the values of x, y, and z.

The equation of the line is given as:

x = y - 1/2 = z - 1/2 = 1/2

To find the coordinates at a distance of √11 units from the origin, we can use the distance formula in 3D space:

Distance from the origin = √(x^2 + y^2 + z^2)

Since the distance is √11, we have:

√(x^2 + y^2 + z^2) = √11

Squaring both sides to eliminate the square root:

x^2 + y^2 + z^2 = 11

Now, we can substitute the expressions for x, y, and z from the given equation of the line:

x = 1/2

y - 1/2 = 1/2

z - 1/2 = 1/2

From the second equation, we can isolate y:

y = 1/2 + 1/2 = 1

From the third equation, we can isolate z:

z = 1/2 + 1/2 = 1

Now, we can find the values of x, y, and z:

x = 1/2

y = 1

z = 1

Therefore, the coordinates of the points on the line x/1 = y-1/2 = z-1/2 = 1/2 that are at a distance of √11 units from the origin are (1/2, 1, 1).

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Answer :

• Coordinates of point on the line : (1, 3, 1), (-1, -1, -3).

Given :

• x/1 = y - 1/2 = z + 1/2

To Find :

• The coordinates of point line ?

Solution :

• Let assume that, given digits are x/1 = y - 1/2 = z + 1/2 = k

So,

• Point on the line say P is given by x = k, y = 2k + 1, z = 2k - 1

So, given that

• Coordinates of P is (k, 2k + 1, 2k - 1)

Now,

• It is given that, the point P is at a distance of √11 units from origin.

➛ OP = √11

• On squaring both sides, we get

➛ (OP)² = 11

➛ (k - 0)² + (2k + 1 - 0)² + (2k - 1 - 0)² = 11

➛ (k)² + (2k + 1)² + (2k - 1)² = 11

➛ k² + 4k² + 1 + 4k + 4k² + 1 - 4k = 11

➛ 9k² + 2 = 11

➛ 9k² = 11 - 2

➛ 9k² = 9

➛ k² = 1

➛ k = ± 1

We get,

Coordinates of P —

• {(1, 3, 1), when k = 1}
• (-1, -1, -3), when k = -1

Hence,

• Coordinates of point on the line : (1, 3, 1), (-1, -1, -3).