Consider the line  ℓ:r(t)=(2,5,3)+t(9,6,7),t∈R
 . Let  P(x,y,z)
  be the point on the line  ℓ
  that is nearest to the origin. Find  x,y
  and  z
 . Express your solution in exact form.

 

To find the point \( P(x, y, z) \) on the line \(\ell\) that is nearest to the origin, we can use the fact that the vector from the origin to \( P \) must be perpendicular to the direction vector of the line. Here's the step-by-step solution:

### Step 1: Parametric Equation of the Line
The line \(\ell\) is given by the parametric equation:
\[
r(t) = (2, 5, 3) + t(9, 6, 7)
\]
This can be written as:
\[
r(t) = (2 + 9t, 5 + 6t, 3 + 7t)
\]
So the coordinates of any point \( P \) on the line are:
\[
P(t) = (x, y, z) = (2 + 9t, 5 + 6t, 3 + 7t)
\]

### Step 2: Vector from the Origin to \( P(t) \)
The vector from the origin to the point \( P(t) \) is:
\[
\mathbf{OP} = \left(2 + 9t, 5 + 6t, 3 + 7t\right)
\]

### Step 3: Perpendicularity Condition
The vector \(\mathbf{OP}\) should be perpendicular to the direction vector of the line \( \mathbf{d} = (9, 6, 7) \). For perpendicularity, their dot product must be zero:
\[
\mathbf{OP} \cdot \mathbf{d} = 0
\]
Substitute the vectors:
\[
(2 + 9t, 5 + 6t, 3 + 7t) \cdot (9, 6, 7) = 0
\]

### Step 4: Compute the Dot Product
Compute the dot product:
\[
(2 + 9t) \cdot 9 + (5 + 6t) \cdot 6 + (3 + 7t) \cdot 7 = 0
\]
Simplify:
\[
18 + 81t + 30 + 36t + 21 + 49t = 0
\]
\[
81t + 36t + 49t + 18 + 30 + 21 = 0
\]
\[
166t + 69 = 0
\]

### Step 5: Solve for \( t \)
Solve for \( t \):
\[
t = -\frac{69}{166}
\]

### Step 6: Find the Coordinates \( x, y, z \)
Now substitute \( t = -\frac{69}{166} \) back into the parametric equations:
\[
x = 2 + 9\left(-\frac{69}{166}\right) = 2 - \frac{621}{166} = \frac{332 - 621}{166} = -\frac{289}{166}
\]
\[
y = 5 + 6\left(-\frac{69}{166}\right) = 5 - \frac{414}{166} = \frac{830 - 414}{166} = \frac{416}{166} = \frac{208}{83}
\]
\[
z = 3 + 7\left(-\frac{69}{166}\right) = 3 - \frac{483}{166} = \frac{498 - 483}{166} = \frac{15}{166}
\]

### Final Answer:
The coordinates of the point \( P(x, y, z) \) on the line \(\ell\) nearest to the origin are:
\[
x = -\frac{289}{166}, \quad y = \frac{208}{83}, \quad z = \frac{15}{166}
\]

Would you like more details on any specific part, or do you have any questions?

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Here are some related questions:
1. How do you determine the distance from a point to a line in 3D space?
2. What is the significance of the perpendicularity condition in finding the nearest point?
3. How would the solution change if the line was expressed differently?
4. How can the dot product be used to determine orthogonality in other contexts?
5. What are other methods to find the closest point between two skew lines?

**Tip:** When dealing with problems in three-dimensional geometry, visualizing the situation can greatly help in understanding the relationships between vectors and points.

Consider the line ℓ: r(t) = (2,5,3) + t(9,6,7), t ∈ R. Let P(x,y,z) be the point on the line ℓ that is nearest to the origin. Find x, y, and z. Express your solution in exact form.

Discover how to determine the point on the line r(t) = (2,5,3) + t(9,6,7) that is closest to the origin using vector geometry and the dot product. This problem involves finding the exact coordinates of the point and covers essential concepts like the parametric equations of a line and the perpendicularity condition. Ideal for students in Grades 11-12.

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