The distance $d$ of a point $P$ to the line through points $A$ and $B$ is the length of the component of $\overline{A P}$ that is orthogonal to $\overline{A B}$, as indicated in the diagram.

So the distance from $P=(2,5,0)$ to the line through the points $A=(0,3,-3)$ and $B=(-3,-5,0)$ is

Question

The distance $d$ of a point $P$ to the line through points $A$ and $B$ is the length of the component of $\overline{A P}$ that is orthogonal to $\overline{A B}$, as indicated in the diagram.

So the distance from $P=(2,5,0)$ to the line through the points $A=(0,3,-3)$ and $B=(-3,-5,0)$ is

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the distance from point $P = (2, 5, 0)$ to the line through points $A = (0, 3, -3)$ and $B = (-3, -5, 0)$, we first need to determine the direction vector of the line $AB$. The direction vector $\overrightarrow{AB}$ is calculated as follows: $$ \overrightarrow{AB} = B - A = (-3 - 0, -5 - 3, 0 + 3) = (-3, -8, 3) $$ ‖ ‖ step 2 ⋮ Next, we find the vector $\overrightarrow{AP}$ from point $A$ to point $P$: $$ \overrightarrow{AP} = P - A = (2 - 0, 5 - 3, 0 + 3) = (2, 2, 3) $$ ‖ ‖ step 3 ⋮ We now need to project $\overrightarrow{AP}$ onto $\overrightarrow{AB}$. The projection formula is: $$ \text{proj}_{\overrightarrow{AB}} \overrightarrow{AP} = \frac{\overrightarrow{AP} \cdot \overrightarrow{AB}}{\overrightarrow{AB} \cdot \overrightarrow{AB}} \overrightarrow{AB} $$ First, calculate the dot product $\overrightarrow{AP} \cdot \overrightarrow{AB}$: $$ \overrightarrow{AP} \cdot \overrightarrow{AB} = (2)(-3) + (2)(-8) + (3)(3) = -6 - 16 + 9 = -13 $$ Next, calculate $\overrightarrow{AB} \cdot \overrightarrow{AB}$: $$ \overrightarrow{AB} \cdot \overrightarrow{AB} = (-3)^2 + (-8)^2 + (3)^2 = 9 + 64 + 9 = 82 $$ Thus, the projection is: $$ \text{proj}_{\overrightarrow{AB}} \overrightarrow{AP} = \frac{-13}{82} \overrightarrow{AB} = \frac{-13}{82} (-3, -8, 3) = \left( \frac{39}{82}, \frac{104}{82}, \frac{-39}{82} \right) $$ ‖ ‖ step 4 ⋮ The orthogonal component of $\overrightarrow{AP}$ to $\overrightarrow{AB}$ is: $$ \overrightarrow{AP} - \text{proj}_{\overrightarrow{AB}} \overrightarrow{AP} = \left( 2, 2, 3 \right) - \left( \frac{39}{82}, \frac{104}{82}, \frac{-39}{82} \right) = \left( 2 - \frac{39}{82}, 2 - \frac{104}{82}, 3 + \frac{39}{82} \right) = \left( \frac{123}{82}, \frac{60}{82}, \frac{285}{82} \right) $$ ‖ ‖ step 5 ⋮ Finally, the distance $d$ is the magnitude of this orthogonal component: $$ d = \left| \left| \left( \frac{123}{82}, \frac{60}{82}, \frac{285}{82} \right) \right| \right| = \sqrt{ \left( \frac{123}{82} \right)^2 + \left( \frac{60}{82} \right)^2 + \left( \frac{285}{82} \right)^2 } = \sqrt{ \frac{15129}{6724} + \frac{3600}{6724} + \frac{81225}{6724} } = \sqrt{ \frac{99954}{6724} } = \frac{\sqrt{99954}}{82} = \frac{315.97}{82} \approx 3.8651 $$ ‖ ∻Answer∻ ⚹ The distance from point $P = (2, 5, 0)$ to the line through points $A = (0, 3, -3)$ and $B = (-3, -5, 0)$ is approximately $3.8651$. ⚹ ∻Key Concept∻ ⚹Distance from a point to a line in 3D space⚹ ∻Explanation∻ ⚹The distance from a point to a line in 3D space is found by projecting the vector from the point to a point on the line onto the direction vector of the line, and then finding the orthogonal component of this projection. The magnitude of this orthogonal component gives the distance.⚹◊What is the formula for calculating the distance between a point and a line in 3D space?⍭ Generate me a similar question◊

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