Distance Between Points A And B On Line L

Point A and Line L: Finding the Distance Between A and B

Hey guys, let's dive into a cool math problem today that involves points, lines, and a bit of geometric wizardry. We're going to figure out the distance between a specific point and a line, but with a twist! We've got Point $A$ chilling at coordinates $(7,-1,8)$. Then there's this line, $L$, which is defined by the Cartesian equation rac{x-3}{1}=rac{y+1}{2}=z. Our mission, should we choose to accept it, is to find a point, let's call it $B$, that sits on line $L$. The catch is that the line segment $A B$ has to be perpendicular to line $L$. Once we find that special point $B$, we'll calculate the distance between $A$ and $B$. This is a classic problem that tests our understanding of vector geometry and how to find the shortest distance from a point to a line in 3D space.

Understanding the Geometry of the Problem

First off, let's break down what we're dealing with here. We have a point $A$ in 3D space, given by its coordinates $(7,-1,8)$. Think of it as a specific location in our universe. Then we have a line $L$, described by a set of symmetric equations: rac{x-3}{1}=rac{y+1}{2}=z. This tells us how to navigate along the line. Any point on this line can be represented using a parameter, say $t$. From the equations, we can see that $x = 3 + t$, $y = -1 + 2t$, and $z = 0 + t$. So, any point on line $L$ can be written in parametric form as $P(t) = (3+t, -1+2t, t)$. Our goal is to find a specific point $B$ on this line $L$. This means $B$ will have coordinates of the form $(3+t_B, -1+2t_B, t_B)$ for some value of $t_B$. The crucial condition is that the line segment $A B$ must be perpendicular to line $L$. In vector terms, this means the vector vec{AB} must be orthogonal to the direction vector of line $L$. The direction vector of line $L$ can be read directly from the denominators of the symmetric equations, which is vec{d} = egin{pmatrix} 1 \ 2 \ 1 mathend{pmatrix}. The vector vec{AB} connects point $A$ to point $B$. If $A = (x_A, y_A, z_A)$ and $B = (x_B, y_B, z_B)$, then vec{AB} = egin{pmatrix} x_B - x_A \ y_B - y_A \ z_B - z_A mathend{pmatrix}. Since $B$ lies on $L$, its coordinates are $(3+t_B, -1+2t_B, t_B)$. Therefore, vec{AB} = egin{pmatrix} (3+t_B) - 7 \ (-1+2t_B) - (-1) \ t_B - 8 mathend{pmatrix} = egin{pmatrix} t_B - 4 \ 2t_B \ t_B - 8 mathend{pmatrix}. The condition of perpendicularity means the dot product of vec{AB} and the direction vector vec{d} is zero: vec{AB} cdot vec{d} = 0. This will give us an equation to solve for $t_B$. Once we find $t_B$, we can find the coordinates of $B$ and then calculate the distance $AB$. This approach leverages the fundamental properties of vectors and lines in three-dimensional Euclidean space, making it a solid method for solving this type of problem. The process involves setting up vector equations and solving for unknown parameters, which is a common theme in analytical geometry.

Setting Up the Vector Equations

Alright, let's get down to business with the vector equations. We know Point $A$ is $(7,-1,8)$. Line $L$ can be parameterized. A general point $P$ on line $L$ can be represented as P(t) = A_0 + tvec{d}, where $A_0$ is a point on the line and vec{d} is the direction vector. From the equation rac{x-3}{1}=rac{y+1}{2}=z, we can identify a point on the line by setting the numerators to zero, which gives us $x=3$, $y=-1$, and $z=0$. So, $A_0 = (3,-1,0)$ is a point on $L$. The direction vector vec{d} is given by the denominators: vec{d} = egin{pmatrix} 1 \ 2 \ 1 mathend{pmatrix}. Thus, any point on line $L$ can be written as P(t) = (3,-1,0) + tegin{pmatrix} 1 \ 2 \ 1 mathend{pmatrix} = (3+t, -1+2t, t). Now, let Point $B$ be a point on line $L$. This means its coordinates must be of the form $(3+t_B, -1+2t_B, t_B)$ for some specific parameter value $t_B$. We are given that the line segment $A B$ is perpendicular to line $L$. This means the vector vec{AB} is orthogonal to the direction vector vec{d} of line $L$. Let's form the vector vec{AB}. The coordinates of $A$ are $(7,-1,8)$ and the coordinates of $B$ are $(3+t_B, -1+2t_B, t_B)$. So, vec{AB} = B - A = egin{pmatrix} (3+t_B) - 7 \ (-1+2t_B) - (-1) \ t_B - 8 mathend{pmatrix} = egin{pmatrix} t_B - 4 \ 2t_B \ t_B - 8 mathend{pmatrix}. The condition for perpendicularity is that the dot product of vec{AB} and vec{d} is zero: vec{AB} cdot vec{d} = 0. Substituting the components, we get: egin{pmatrix} t_B - 4 \ 2t_B \ t_B - 8 mathend{pmatrix} cdot egin{pmatrix} 1 \ 2 \ 1 mathend{pmatrix} = 0. This expands to $(t_B - 4)(1) + (2t_B)(2) + (t_B - 8)(1) = 0$. This equation will allow us to solve for the specific value of $t_B$ that defines point $B$. Once we have $t_B$, we can determine the exact coordinates of $B$ and then proceed to calculate the distance between $A$ and $B$. This step is crucial as it directly uses the geometric constraint given in the problem statement to find the location of point B.

Solving for Point B's Coordinates

Let's take the equation we derived from the dot product and solve it for $t_B$. Remember, this equation represents the condition that the vector vec{AB} is perpendicular to the direction vector of line $L$. The equation is: $(t_B - 4)(1) + (2t_B)(2) + (t_B - 8)(1) = 0$. Let's simplify and solve this linear equation. First, distribute the terms: $t_B - 4 + 4t_B + t_B - 8 = 0$. Now, combine the terms involving $t_B$: $(1+4+1)t_B = 6t_B$. Combine the constant terms: $-4 - 8 = -12$. So, the equation becomes: $6t_B - 12 = 0$. Now, we can solve for $t_B$. Add 12 to both sides: $6t_B = 12$. Finally, divide by 6: t_B = rac{12}{6} = 2. So, the parameter value for point $B$ on line $L$ is $t_B = 2$. Now that we have $t_B$, we can find the exact coordinates of point $B$. Recall that points on line $L$ have the form $(3+t, -1+2t, t)$. Substituting $t_B = 2$, we get: $x_B = 3 + 2 = 5$. $y_B = -1 + 2(2) = -1 + 4 = 3$. $z_B = 2$. Therefore, the coordinates of point $B$ are $(5,3,2)$. We've successfully located point $B$ on line $L$ such that $AB$ is perpendicular to $L$. This is a significant step, as we've used the vector orthogonality condition to pin down the exact location of $B$. This process confirms our understanding of how to apply vector algebra to solve geometric problems in 3D space. Finding these coordinates is key to the next step, which is calculating the distance.

Calculating the Distance Between A and B

Now that we've found the coordinates of point $B$ to be $(5,3,2)$, and we know the coordinates of point $A$ are $(7,-1,8)$, we can finally calculate the distance between these two points. The distance formula in three dimensions between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by d = sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}. Let's plug in the coordinates of $A$ and $B$: $x_A=7, y_A=-1, z_A=8$ and $x_B=5, y_B=3, z_B=2$. The distance $AB$ is: AB = sqrt{(5-7)^2 + (3-(-1))^2 + (2-8)^2}. Let's calculate the differences: $5-7 = -2$. $3-(-1) = 3+1 = 4$. $2-8 = -6$. Now, square these differences: $(-2)^2 = 4$. $(4)^2 = 16$. $(-6)^2 = 36$. Add the squared differences: $4 + 16 + 36 = 56$. So, the distance AB = sqrt{56}. We can simplify sqrt{56} by finding any perfect square factors of 56. $56 = 4 imes 14$. So, sqrt{56} = sqrt{4 imes 14} = sqrt{4} imes sqrt{14} = 2sqrt{14}. Therefore, the distance between points $A$ and $B$ is 2sqrt{14}. This is the final answer, representing the shortest distance from point $A$ to the line $L$, as the line segment $AB$ is perpendicular to $L$. This calculation completes our problem, showcasing how to use vector projections and the distance formula in tandem to solve geometric challenges in 3D space. It's a satisfying conclusion to a problem that started with just a point and a line definition!

Conclusion: The Shortest Distance Found

So there you have it, guys! We've successfully navigated the world of 3D coordinates and vector geometry to find the distance between Point $A$ and Point $B$. We started with Point $A$ at $(7,-1,8)$ and a line $L$ defined by rac{x-3}{1}=rac{y+1}{2}=z. Our goal was to find a point $B$ on $L$ such that the line segment $A B$ is perpendicular to $L$. By parameterizing the line $L$ and forming the vector vec{AB}, we used the condition of perpendicularity (the dot product being zero) to solve for the parameter $t_B$. This gave us $t_B = 2$, which led us to the coordinates of $B$ as $(5,3,2)$. Finally, we applied the 3D distance formula to find the length of the segment $A B$, which turned out to be sqrt{56}, or 2sqrt{14}. This distance, 2sqrt{14}, is not just any distance; it's the shortest distance from Point $A$ to the line $L$. This method is a fundamental technique in vector calculus and analytical geometry for solving problems involving distances and perpendicularity in three dimensions. It's a testament to how powerful vector algebra is in simplifying complex geometric relationships. Keep practicing these problems, and you'll become a pro at spotting these geometric connections in no time! It's all about setting up the right equations based on the geometric properties you're given. Awesome job, everyone!