Question

Line \(\underline{r}=(1,2,1)+\mathrm{k}(-1,-2,1), \mathrm{k} \in \mathrm{R}\) the point which is at \(\sqrt{6}\) dist. away from \((2,4,0)\) is (A) \((1,2,1)(3,6,-1)\) (B) \((1,2,1)(3,-6,-1)\) (C) \((-1,-2,1)(3,6,-1)\) (D) None of these

Short answer

The point on the line which is at a distance of \(\sqrt{6}\) away from (2, 4, 0) is (D) None of these since the quadratic equation for k produced no real solutions.

Step-by-step solution

Write down the parametric equations of the line.

Parametric equations of the line \(\underline{r}\) are given by: \(x = 1 - k\) \(y = 2 - 2k\) \(z = 1 + k\) where \(\mathrm{k} \in \mathrm{R}\).

Use the distance formula to write down the equation for the distance between two points.

We need to find a point on the given line with a distance of \(\sqrt{6}\) from the point (2, 4, 0). We can use the distance formula to achieve this. The distance formula for points in 3D-space is: \(D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\) where (x_1, y_1, z_1) and (x_2, y_2, z_2) are the coordinates of two points in the 3D-space.

Substitute the coordinates of the points and the value of the desired distance into the distance formula.

We know the desired distance \(D = \sqrt{6}\) and the point (2, 4, 0). We can substitute these values into the distance formula. Since we are looking for a point on the line, we can substitute the parametric equations of the line for (x_2, y_2, z_2). \(\sqrt{6} = \sqrt{((1 - k) - 2)^2 + ((2-2k) - 4)^2 + ((1+k) - 0)^2}\)

Solve for k.

Now, we need to find the value of k that satisfies the equation. We can square both sides of the equation to get rid of the square root. \(6 = (3 - k)^2 + (-2-2k)^2 + (1+k)^2\) Now, we can expand the equation: \(6 = 9 - 6k + k^2 + 4 + 8k + 4k^2 + 1 + 2k + k^2\) Combining the terms, we get: \(6 = 14 + 4k + 6k^2\) Subtract 6 from both sides: \(0 = 8 + 4k + 6k^2\) Now, we need to solve this quadratic equation to find the value(s) of k.

Solve the quadratic equation for k.

We can divide the equation by 2 to simplify it: \(0 = 4 + 2k + 3k^2\) Now, rearrange the quadratic equation to standard form: \(0 = 3k^2 + 2k + 4\) We can't factor the equation easily, so let's use the quadratic formula: \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Here, a = 3, b = 2, and c = 4. Applying the formula: \(k = \frac{-2 \pm \sqrt{2^2 - 4(3)(4)}}{2(3)}\) Since the discriminant is negative, there are no real solutions for k. Therefore, we conclude that the answer is: (D) None of these

Key concepts

Distance Formula

The Distance Formula is a powerful tool for finding the distance between two points in a 3D space. When dealing with points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), the formula takes the form: \[D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]This equation, derived from the Pythagorean theorem, helps us measure how far apart two points are in a three-dimensional coordinate system.

• Subtract the coordinates of the first point from the corresponding coordinates of the second point.
• Square each difference.
• Add the three squared differences together.
• Take the square root of the resulting sum to find the distance.

Using this formula allows us to calculate precise distances, essential for tasks such as finding points on lines or determining spatial relationships.

3D Coordinate Geometry

3D Coordinate Geometry extends the principles of 2D geometry into three dimensions, introducing the z-axis to the usual x and y axes. This addition provides a framework to work with spatial objects, lines, and curves in a 3D space.In 3D coordinate geometry:

• A point is represented as \((x, y, z)\).
• A line can be expressed using parametric equations that depend on a parameter, often denoted by \(k\).

For example, a line's equation might look like:\[x = 1 - k\y = 2 - 2k\z = 1 + k\]Here, as \(k\) varies, the point \((x, y, z)\) moves along the line.This method of expressing lines is particularly useful for calculations involving distances and intersections because it provides a flexible representation of line segments in 3D.

Quadratic Formula

The Quadratic Formula is a solution to quadratic equations of the form \(ax^2 + bx + c = 0\). It provides a straightforward way to find the roots of these equations even if they cannot be easily factored. The formula is:\[k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• "\(a\)" represents the coefficient of \(k^2\).
• "\(b\)" is the coefficient of \(k\).
• "\(c\)" is the constant term.

When you substitute the appropriate values into the formula:- Calculate \(b^2 - 4ac\), known as the discriminant.- If the discriminant is positive, two real solutions exist.- If zero, one real solution.- If negative, no real solutions.The Quadratic Formula is crucial in determining the possible values of parameters (like \(k\) in our problem) when solving geometric problems in 3D.