Question

Task 33 - Lines and Points Σ = 45 points Let's consider the points P₁ (2; 9; -4) and = P₂ = (-4; -2; 8). (a) What is the equation g of the line through P₁ and P₂? Pay attention to correct notation (vector notation)! Show intermediate steps! (11P) (b) What are the coordinates of the point Q at 40% of the distance P₁ P₂? How can the obtained vector be written using integers? (c) How far is the point R = (3; 2; 1) from? How does R relate to the direction vector of the line? Is it on the same side or the opposite side of point P₁? How can this be understood from the formula and visually (sketch)? The sketch should show all components of the distance formula. (22P)​

Answer 1

the equation of the line
`\(g\)` is
`\[ x = 2 - 6t, \quad y = 9 - 11t, \quad z = -4 + 12t \]`. The coordinates of point (Q) are
`\((-0.4, 4.6, 0.8)\)`. Since
`\((\mathbf{R} - \mathbf{P_0}) \cdot \mathbf{v} = 131\)` (positive), (R) is on the same side.

(a) Equation of the Line:

The equation of the line
`\(g\)` passing through points
`\(P_1(2, 9, -4)\)` and
`\(P_2(-4, -2, 8)\)` can be represented in vector form as follows:

`\[ \mathbf{r} = \mathbf{P_1} + t(\mathbf{P_2} - \mathbf{P_1}) \]`

Where:

-
`\(\mathbf{r}\)` is a vector representing any point on the line,

-
`\(\mathbf{P_1}\)` is the position vector of point
`\(P_1\)`,

-
`\(\mathbf{P_2}\)` is the position vector of point
`\(P_2\)`, and

-
`\(t\)` is a scalar parameter.

Let's calculate the direction vector
`\(\mathbf{P_2} - \mathbf{P_1}\)`:

`\[ \mathbf{P_2} - \mathbf{P_1} = \langle -4 - 2, -2 - 9, 8 - (-4) \rangle = \langle -6, -11, 12 \rangle \]`

Now, the equation of the line is:

`\[ \mathbf{r} = \langle 2, 9, -4 \rangle + t \langle -6, -11, 12 \rangle \]`

So, the equation of the line
`\(g\)` is:

`\[ x = 2 - 6t, \quad y = 9 - 11t, \quad z = -4 + 12t \]`

(b) Coordinates of Point Q at 40% of the Distance:

To find the point
`\(Q\)` at 40% of the distance from
`\(P_1\) to \(P_2\)`, we can use the parameter
`\(t = 0.4\)`:

`\[ \mathbf{Q} = \mathbf{P_1} + 0.4 (\mathbf{P_2} - \mathbf{P_1}) \]`

Calculate
`\(\mathbf{Q}\)`:

`\[ \mathbf{Q} = \langle 2, 9, -4 \rangle + 0.4 \langle -6, -11, 12 \rangle \]`

`\[ \mathbf{Q} = \langle 2, 9, -4 \rangle + \langle -2.4, -4.4, 4.8 \rangle \]`

`\[ \mathbf{Q} = \langle -0.4, 4.6, 0.8 \rangle \]`

So, the coordinates of point (Q) are
`\((-0.4, 4.6, 0.8)\)`.

(c) Distance from Point R to the Line:

The distance
`(\(d\))` between a point
`\(R(x_0, y_0, z_0)\)` and a line with direction vector
`\(\mathbf{v} = \langle a, b, c \rangle\)` and passing through the point
`\(\mathbf{P_0} = \langle x_1, y_1, z_1 \rangle\)` is given by the formula:

`\[ d = \frac{\left|(\mathbf{R} - \mathbf{P_0}) \cdot \mathbf{v}\right|}{\|\mathbf{v}\|} \]`

Here,
`\(\cdot\)` represents the dot product, and
`\(\|\mathbf{v}\|\)` is the magnitude of
`\(\mathbf{v}\).`

Let
`\(R(3, 2, 1)\)` be the point, and
`\(\mathbf{v} = \langle -6, -11, 12 \rangle\)` be the direction vector. Also, let
`\(\mathbf{P_0} = \langle 2, 9, -4 \rangle\)`.

Calculate the distance:

`\[ \mathbf{R} - \mathbf{P_0} = \langle 3 - 2, 2 - 9, 1 - (-4) \rangle = \langle 1, -7, 5 \rangle \]`

`\[ \|\mathbf{v}\| = √((-6)^2 + (-11)^2 + 12^2) = √(36 + 121 + 144) = √(301) \]`

`\[ \left|(\mathbf{R} - \mathbf{P_0}) \cdot \mathbf{v}\right| = | \langle 1, -7, 5 \rangle \cdot \langle -6, -11, 12 \rangle | = |-6 + 77 + 60| = 131 \]`

`\[ d = (131)/(√(301)) \]`

So, the distance (d) is given by
`\(d = (131)/(√(301))\)`. To determine whether (R) is on the same side or the opposite side of
`\(P_1\)`, we can check the sign of
`\((\mathbf{R} - \mathbf{P_0}) \cdot \mathbf{v}\)`.

If it's positive, (R) is on the same side; if negative, it's on the opposite side. In this case, since
`\((\mathbf{R} - \mathbf{P_0}) \cdot \mathbf{v} = 131\)` (positive), (R) is on the same side.