A Mathematical Blog

In this short post, I will be showing a three step process to constructing the Centroid in Geogebra - an online graphing calculator.  Centroid is the point where the medians of a triangle intersect. A median is a segment connecting a vertex of a triangle to the midpoint on the opposite side.  Read more  : Medians and the Centroid of a Triangle Step 1 : Constructing a triangle. For the sake of this tutorial, I will be constructing any arbitrary triangle using the Polygon tool.  Polygon tool is located under the Triangles section in the shapes menu.  Select this tool simply by a left mouse click, or a single tap on screen. The name of the selected tool will appear as bold to indicate that the tool has been selected.  On the graphing panel to the right, click on any three spots to mark the locations of the three vertices of your triangle. Click on the first vertex again to complete the construction of a three sided triangle.  To know about how to construct a triangle of three kno

Two vectors are perpendicular if their directions are 90 degrees apart. The dot product of two perpendicular vectors is zero.  A Normal vector to a plane A normal vector to a plane is basically a vector that makes a ninety degree angle with the plane(perpendicular to the plane).   In the above picture, \(\vec{d}\) is a normal vector to the plane, P. A normal vector to the plane will be perpendicular to every other vector which is contained in the plane or parallel to the plane.  Equation of a plane To find the equation of a plane we need : Any non-zero vector normal to the plane Coordinates of any point in the plane Consider a plane, P. Let there be a point A(-1, -1, 1) in the plane whose coordinates are known to us. Let point B(x, y, z) be an arbitrary point in the plane. It is an arbitrary point, so it could be anywhere on the plane.  Let, \(\overrightarrow{d} = -2\hat{i} + 2\hat{j} + 0\hat{k}\) be a normal vector to the plane. The picture below roughly depicts how every

In this post I will be talking about how to find vector equation of a line in 3D.  To find vector equation of a line we need : A direction vector of the line Coordinates of any point on the line Equation from direction vector and a point on the line D irection or directional vector is a vector arrow parallel to the line. Some texts will use the term ‘Direction ratios’ or ‘Direction Numbers’ for direction vector. Direction numbers or direction ratios of a line is a set of three numbers that specify the direction of that line. A line with direction ratios of 1, 2, 3 basically means that \(\hat{i} + \hat{j} + \hat{k}\) is its direction vector, i.e. a vector parallel to that line.  The procedure for finding the equation of a line is simple enough. Consider a line in 3D space; let’s name it L.  Suppose \(\vec{d} = 3\hat{i} + 2\hat{j} - 8\hat{k}\) is its direction vector.  Let P(5, 2, -4) be a point on L. Below is a rough picture of how everything is supposed to look like.  Defin

In my previous post on the topic, i clarified that the shortest path between a pair of skew lines is the line segment that is perpendicular to both the skew lines.  Now the next ask is determining a way to find the length of the shortest path, i.e. the length of the line segment perpendicular to both the skew lines. Let’s take an example of a pair of skew lines. L1 : \(\hat{i} + \hat{j} + 0\hat{k} + u(2\hat{i} - \hat{j} + \hat{k})\) L2 : \(2\hat{i} + \hat{j} - \hat{k} + v(3\hat{i} - 5\hat{j} + 2\hat{k})\) From equation of L1, position and direction vectors of L1 are  \(\overrightarrow{p_1} = \hat{i} + \hat{j} + 0\hat{k}\) \(\overrightarrow{d_1} = 2\hat{i} - \hat{j} + \hat{k}\) From equation of L2, position and direction vectors of L2 are \(\overrightarrow{p_2} = 2\hat{i} + \hat{j} - \hat{k}\) \(\overrightarrow{d_2} = 3\hat{i} - 5\hat{j} + 2\hat{k}\) Direction vector of a line is simply a vector arrow parallel to that line. Or, we can also say it is a vector arrow pointing