plane line intersection

This means that every value of \(t\) will produce a point on the line that is also on the plane, telling us that the line is contained in the plane whose equation is \( x + 2y - 2z = -1\). Sign in to comment. So the point of intersection of this line with this plane is \(\left(5, -2, -9\right)\). Those values generate the points on the two segments that are closest to the point of intersection. If we take the parameter at being one of the coordinates, this usually simplifies the algebra. How can we differentiate between these three possibilities? Or the line could completely lie inside the plane. Intersection of plane and line. The line case is a lot easier because any two non-parallel lines in an x,y plane will intersect somewhere, not so with segments – user316117 Dec 28 '15 at 18:31 Do you mean lines or line segments? 0. Therefore, coordinates of intersection must satisfy both equations, of the line and the plane. For and , this means that all ratios have the value a, or that for all i. Notice that we can substitute the expressions of \(t\) given in the parametric equations of the line into the plane equation for \(x\), \(y\), and \(z\). As shown in the diagram above, two planes intersect in a line. 0 ⋮ Vote. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), [ "article:topic", "authorname:pseeburger", "license:ccby" ], \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\). Watch the recordings here on Youtube! Vote. Have questions or comments? You say "lines" but you say they have length. How can we tell if a line is contained in the plane? Example \(\PageIndex{9}\): Other relationships between a line and a plane, \[\begin{align*} \text{Line:}\quad x &=1 + 2t & \text{Plane:} \quad x + 2y - 2z = 5 \\[5pt] y &= -2 + 3t \\[5pt] z &= -1 + 4t \end{align*}\nonumber\]. If they do intersect, determine whether the line is contained in the plane or intersects it in a single point. Collecting like terms on the left side causes the variable \(t\) to cancel out and leaves us with a contradiction: Since this is not true, we know that there is no value of \(t\) that makes this equation true, and thus there is no value of \(t\) that will give us a point on the line that is also on the plane. Example: Find the equation of intersection of the planesand, We take the parameter asand putThe equations become, Finding the Line of Intersection of Two Planes, The Image of a Line Under a Transformation Represented by a Matrix, Constructions - Bisecting Angles and Lines - Constructing an Angle of 60 Degrees, Constructing a Set of Points a Fixed Distance From a Given Line. Line-Plane Intersection The plane determined by the points,, and and the line passing through the points and intersect in a point which can be determined by … If we take the parameter at being one of the coordinates, this usually simplifies the algebra. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Follow 40 views (last 30 days) Stephanie Ciobanu on 9 Nov 2017. Solution: Because the intersection point is common to the line and plane we can substitute the line parametric points into the plane equation to get: 4 (− 1 − 2t) + (1 + t) − 2 = 0. t = − 5/7 = 0.71. Finally, if the line intersects the plane in a single point, determine this point of intersection. This is equivalent to the conditions that all . If we take the parameter at being one of the coordinates, this usually simplifies the algebra. The code then adjusts t1 and t2 so they are between 0 and 1. Determine whether the following line intersects with the given plane. We can verify this by putting the coordinates of this point into the plane equation and checking to see that it is satisfied. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. In Euclidean geometry, the intersection of a line and a line can be the empty set, a point, or a line. Sign in to answer this question. Example \(\PageIndex{8}\): Finding the intersection of a Line and a plane. Finally the code uses the adjusted values of t1 and t2 to find those closest points. We can find the equation of the line by solving the equations of the planes simultaneously, with one extra complication – we have to introduce a parameter. Line-Plane Intersection The plane determined by the points , , and and the line passing through the points and intersect in a point which can be determined by … Show Hide all comments. As shown in the diagram above, two planes intersect in a line. Here you can calculate the intersection of a line and a plane (if it exists). Finally, if the line intersects the plane in a single point, determine this point of intersection. There are three possibilities: The line could intersect the plane in a point. There are three possibilities: The line could intersect the plane in a point. Line: x = 2 − t Plane: 3 x − 2 y + z = 10 y = 1 + t z = 3 t. Now that we have examined what happens when there is a single point of intersection between a line and a point, let's consider how we know if the line either does not intersect the plane at all or if it lies on the plane (i.e., every point on the line is also on the plane). But the line could also be parallel to the plane. Even if this plane and line is not intersecting, it shows check=1 and intersection point I =[-21.2205 31.6268 6.3689]. Since we found a single value of \(t\) from this process, we know that the line should intersect the plane in a single point, here where \(t = -3\). Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Distinguishing these cases and finding the intersection point have use, for example, in computer graphics, motion planning, and collision detection. Simply enter your exercise and it will be solved step by step. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. But the line could also be parallel to the plane. This is Mathepower. If t1 and t2 are both between 0 and 1, then the line segments intersect. What if we keep the same line, but modify the plane equation to be \( x + 2y - 2z = -1\)? 0 Comments . The point of intersection is a common point of a line and a plane. Now we can substitute the value of t into the line parametric equation to get the intersection point. This means that this line does not intersect with this plane and there will be no point of intersection. A given line and a given plane may or may not intersect. Do a line and a plane always intersect? In 2D, with and , this is the perp prod… As shown in the diagram above, two planes intersect in a line. Check: \(3(5) - 2(-2) + (-9) = 15 + 4 - 9 = 10\quad\checkmark\). No. The parametric equation of a line, x = x0 + at, y = y0 + bt and z = z0 + ct Of course. Can you please explain what is the issue? Missed the LibreFest? These lines are parallel when and only when their directions are collinear, namely when the two vectors and are linearly related as u = av for some real number a. If the line does intersect with the plane, it's possible that the line is completely contained in the plane as well. We can find the equation of the line by solving the equations of the planes simultaneously, with one extra complication – we have to introduce a parameter. So the point of intersection can be determined by plugging this value in for \(t\) in the parametric equations of the line. We can find the equation of the line by solving the equations of the planes simultaneously, with one extra complication – we have to introduce a parameter. Can i see some examples? Substituting the expressions of \(t\) given in the parametric equations of the line into the plane equation gives us: \[(1+2t) +2(-2+3t) - 2(-1 + 4t) = 5\nonumber\]. Legal. In this case, repeating the steps above would again cause the variable \(t\) to be eliminated from the equation, but it would leave us with an identity, \(-1 = -1\), rather than a contradiction. Commented: Star Strider on 9 Nov 2017 Accepted Answer: Star Strider. \[\begin{align*} \text{Line:}\quad x &=2 - t & \text{Plane:} \quad 3x - 2y + z = 10 \\[5pt] y &= 1 + t \\[5pt] z &= 3t \end{align*}\nonumber\]. Here: \(x = 2 - (-3) = 5,\quad y = 1 + (-3) = -2, \,\text{and}\quad z = 3(-3) = -9\). Or the line could completely lie inside the plane. Can i see some examples? what is the code to find the intersection of the plane x + 2y + 3z = 4 and line (x, y, z) = (2,4,6) + t(1,1,1)? If they do intersect, determine whether the line is contained in the plane or intersects it in a single point.

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