Find the distance from the point P(0, \ 2, \ 2) to the line x=2t, \ y= 6 2t, \ z=2+t .

Question:

Find the distance from the point {eq}P(0, \ 2, \ 2) {/eq} to the line {eq}x=2t, \ y= 6-2t, \ z=2+t {/eq}.

Distance between Two Points:

Any point between the 3-dimensional cartesian space can be expressed as {eq}\displaystyle (x, y, z) {/eq}. The distance between any two points can be solved with the equation, {eq}\displaystyle d = \sqrt{(x_0-x_1)^2+(y_0-y_1)^2+(z_0-z_1)^2} {/eq}, where one point is {eq}\displaystyle (x_0,y_0, z_0) {/eq} and the other is {eq}\displaystyle (x_1,y_1,z_1) {/eq}.

Answer and Explanation:

For this problem, we need to find the distance of the vector that is perpendicular from the given line with one end as the given point, P. Any point on the line, S, can be expressed as {eq}\displaystyle S(2t, 6-2t, 2+t) {/eq}. Now, we express the length of any point on the line to P as

{eq}\begin{align} \displaystyle \vec{ SL} &= P - S\\ &= (0-2t, 2-6+2t, 2-2-t)\\ &= (-2t, -4+2t, -t)\\ \end{align} {/eq}

The direction of the line is known as {eq}\displaystyle \vec{L} = (2, -2, 1) {/eq}. Now, we know that {eq}\displaystyle \vec{L} {/eq} and {eq}\displaystyle \vec{SL} {/eq} are perpendicular with each other, so it must be that, {eq}\displaystyle \vec{L}\cdot \vec{SL} = 0 {/eq}. We find t that satisfies the equation.

{eq}\begin{align} \displaystyle \vec{L}\cdot \vec{SL} &= 0\\ (2, -2, 1)\cdot (-2t, -4+2t, -t)&= 0\\ -4+8-4t-t &= 0\\ -5t &= -4\\ t &= \frac{4}{5} \end{align} {/eq}

Therefore, the distance of P to S is the distance between P and the point on the line S at {eq}\displaystyle t = \frac{4}{5} {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle d &= |P -S_{t = \frac{4}{5}}|\\ &= \left| (0, 2, 2) - \left( 2\cdot \frac{4}{5}, 6- 2\cdot \frac{4}{5}, 2 + \frac{4}{5} \right) \right|\\ &= \left|(0,2,2) - \left(\frac{8}{5}, \frac{22}{5}, \frac{14}{5} \right) \right|\\ &= \sqrt{\left(0 - \frac{8}{5} \right)^2-\left(2 - \frac{22}{5} \right)^2-\left(2- \frac{14}{5} \right)^2}\\ &=\boxed{\frac{4\sqrt{14}}{5}} \end{align} {/eq}


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