We are given the line $\ell$ parameterized as:

$r(t) = (6,5,9) + t(7,7,9), \, t \in \mathbb{R}.$

We want to find the point $P(x, y, z)$ on this line that is closest to the origin. The distance from the origin to a point on the line is minimized when the vector from the origin to that point is orthogonal to the direction vector of the line.

Step 1: Parametric equation of the line

The line $\ell$ is parameterized as:

$r(t) = (6, 5, 9) + t(7, 7, 9).$

So, any point $P(x, y, z)$ on the line has coordinates:

$x = 6 + 7t, \quad y = 5 + 7t, \quad z = 9 + 9t.$

Step 2: Vector from the origin to $P(x, y, z)$

The vector from the origin $O(0,0,0)$ to a point $P(x, y, z)$ on the line is given by:

$\vec{OP} = \langle x, y, z \rangle = \langle 6 + 7t, 5 + 7t, 9 + 9t \rangle.$

Step 3: Direction vector of the line

The direction vector of the line $\ell$ is:

$\vec{d} = \langle 7, 7, 9 \rangle.$

Step 4: Condition for the nearest point

The nearest point on the line occurs when the vector $\vec{OP}$ is perpendicular to the direction vector $\vec{d}$. This means that the dot product between $\vec{OP}$ and $\vec{d}$ must be zero:

$\vec{OP} \cdot \vec{d} = 0.$

The dot product is:

$(6 + 7t)(7) + (5 + 7t)(7) + (9 + 9t)(9) = 0.$

Step 5: Solve for $t$

Expanding the terms:

$7(6 + 7t) + 7(5 + 7t) + 9(9 + 9t) = 0,$

$42 + 49t + 35 + 49t + 81 + 81t = 0,$

$158 + 179t = 0.$

Now, solve for $t$:

$179t = -158,$

$t = -\frac{158}{179}.$

Step 6: Find the coordinates $(x, y, z)$

Substitute $t = -\frac{158}{179}$ into the parametric equations for $x$, $y$, and $z$:

$x = 6 + 7\left(-\frac{158}{179}\right) = 6 - \frac{1106}{179} = \frac{1074}{179} - \frac{1106}{179} = \frac{-32}{179},$

$y = 5 + 7\left(-\frac{158}{179}\right) = 5 - \frac{1106}{179} = \frac{895}{179} - \frac{1106}{179} = \frac{-211}{179},$

$z = 9 + 9\left(-\frac{158}{179}\right) = 9 - \frac{1422}{179} = \frac{1611}{179} - \frac{1422}{179} = \frac{189}{179}.$

Thus, the coordinates of the point $P(x, y, z)$ closest to the origin are:

$x = \frac{-32}{179}, \quad y = \frac{-211}{179}, \quad z = \frac{189}{179}.$

Would you like more details or have any questions?

Related Questions:

1. How do you compute the shortest distance between a point and a line in three dimensions?
2. What are the conditions for a vector to be orthogonal to another vector in space?
3. How can the dot product be used to minimize distances in vector calculus problems?
4. How would the solution change if the line was given in symmetric form instead of parametric form?
5. Can you generalize this method for lines in $\mathbb{R}^n$?

Tip:

When finding the closest point on a line, always look for the condition where the vector from the origin (or any point) to the line is perpendicular to the line's direction vector.