Find the point on the line that is closest to the origin.

Question

Find the point on the line  that is closest to the origin.

Quizplus · Accepted Answer

To find the point on the line closest to the origin, we need to minimize the distance between the origin and a point on the line. Let the point on the line be $(x,y,z)$. Then the distance between the origin and this point is given by $\sqrt{x^2+y^2+z^2}$. Now, we want to minimize this subject to the constraint that the point $(x,y,z)$ lies on the line $11eaa3e5_55ba_e025_9f8f_89ea64d0f3c6_\text{TB5971}_\text{11}$. To do this, we use Lagrange multipliers.

Let $f(x,y,z) = \sqrt{x^2+y^2+z^2}$ be the objective function and let $g(x,y,z)=11eaa3e5_55ba_e025_9f8f_89ea64d0f3c6_\text{TB5971}_\text{11}$ be the constraint function. We need to solve the system of equations given by $\nabla f = \lambda \nabla g$ and $g(x,y,z)=11eaa3e5_55ba_e025_9f8f_89ea64d0f3c6_\text{TB5971}_\text{11}$ for $x$, $y$, $z$, and $\lambda$.

Taking partial derivatives, we have:

\begin{align*}
\nabla f &= \frac{1}{\sqrt{x^2+y^2+z^2}}\begin{pmatrix}x\y\z\end{pmatrix}, \
\nabla g &= \begin{pmatrix}3x+4y-2z-15\-2x-y+z+5\x-6y+4z+2\end{pmatrix}.
\end{align*}

Setting these equal and solving, we get:

\begin{align*}
\frac{1}{\sqrt{x^2+y^2+z^2}}\begin{pmatrix}x\y\z\end{pmatrix} &= \lambda \begin{pmatrix}3x+4y-2z-15\-2x-y+z+5\x-6y+4z+2\end{pmatrix}, \
3x+4y-2z-15 &= \lambda x\sqrt{x^2+y^2+z^2}, \
-2x-y+z+5 &= \lambda y\sqrt{x^2+y^2+z^2}, \
x-6y+4z+2 &= \lambda z\sqrt{x^2+y^2+z^2}.
\end{align*}

Squaring the second equation and adding it to four times the first equation and four times the third equation, we get:

$$(3\lambda^2+4\lambda+4)x^2 + (16\lambda^2+4\lambda+4)y^2 + (16\lambda^2-4\lambda+4)z^2 = 225+25\lambda^2.$$

Since this equation must hold for all values of $x$, $y$, and $z$, the coefficients of $x^2$, $y^2$, and $z^2$ must be equal to each other, and we have:

\begin{align*}
3\lambda^2+4\lambda+4 &= 16\lambda^2+4\lambda+4 = 16\lambda^2-4\lambda+4, \
\Rightarrow \lambda &= -\frac{1}{3}, \frac{1}{2}.
\end{align*}

If $\lambda = -\frac{1}{3}$, then we have:

\begin{align*}
3x+4y-2z-15 &= -\frac{1}{3}x\sqrt{x^2+y^2+z^2}, \
-2x-y+z+5 &= -\frac{1}{3}y\sqrt{x^2+y^2+z^2}, \
x-6y+4z+2 &= -\frac{1}{3}z\sqrt{x^2+y^2+z^2}.
\end{align*}

Squaring these equations and adding them, we get:

$$10(x^2+y^2+z^2) = 225.$$

Thus, $x^2+y^2+z^2 = \frac{225}{10} = \frac{45}{2}$, and we also have:

\begin{align*}
9x^2-18xy+12xz &= 186, \
-18xy+16y^2-16yz &= 174, \
12xz-16yz+36z^2 &= 74.
\end{align*}

Solving these equations (for example, by using the first equation to solve for $x^2$, plugging this into the second equation to solve for $y^2$, and so on), we get:

\begin{align*}
x &= \frac{11\sqrt{2}}{10}, \
y &= -\frac{3\sqrt{2}}{5}, \
z &= \frac{3\sqrt{2}}{10}.
\end{align*}

If $\lambda = \frac{1}{2}$, then we have:

\begin{align*}
3x+4y-2z-15 &= \frac{1}{2}x\sqrt{x^2+y^2+z^2}, \
-2x-y+z+5 &= \frac{1}{2}y\sqrt{x^2+y^2+z^2}, \
x-6y+4z+2 &= \frac{1}{2}z\sqrt{x^2+y^2+z^2}.
\end{align*}

Squaring these equations and adding them, we get:

$$35(x^2+y^2+z^2) = 225.$$

Thus, $x^2+y^2+z^2 = \frac{225}{35} = \frac{9}{7}$, and we also have:

\begin{align*}
6x^2-12xy+8xz &= 141, \
-12xy+9y^2-9yz &= 81, \
8xz-9yz+18z^2 &= 38.
\end{align*}

Solving these equations, we get:

\begin{align*}
x &= \frac{6\sqrt{21}}{35}, \
y &= -\frac{3\sqrt{21}}{35}, \
z &= \frac{3\sqrt{21}}{70}.
\end{align*}

Finally, we check which of these points lies on the line $11eaa3e5_55ba_e025_9f8f_89ea64d0f3c6_\text{TB5971}_\text{11}$. Plugging in the coordinates, we find that only choice $\boxed{\textbf{(B) }11eaa3e5_55bb_0737_9f8f_37820c38d0b1_\text{TB5971}_\text{11}}$ satisfies the equation $3x+4y-2z=15$, so this must be the point closest to the origin.

Find the Point on the Line That Is Closest