Commented: Star Strider on 9 Nov 2017 Accepted Answer: Star Strider. 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. Now we can substitute the value of t into the line parametric equation to get the intersection point. As shown in the diagram above, two planes intersect in a line. Finally, if the line intersects the plane in a single point, determine this point of intersection. So the point of intersection of this line with this plane is \(\left(5, -2, -9\right)\). There are three possibilities: The line could intersect the plane in a point. If we take the parameter at being one of the coordinates, this usually simplifies the algebra. This is Mathepower. Simply enter your exercise and it will be solved step by step. Finally, if the line intersects the plane in a single point, determine this point of intersection. \[\begin{align*} \text{Line:}\quad x &=2 - t & \text{Plane:} \quad 3x - 2y + z = 10 \\[5pt] y &= 1 + t \\[5pt] z &= 3t \end{align*}\nonumber\]. Can i see some examples? 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. 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. Of course. How can we differentiate between these three possibilities? As shown in the diagram above, two planes intersect in a line. Vote. If they do intersect, determine whether the line is contained in the plane or intersects it in a single point. Show Hide all comments. But the line could also be parallel to the plane. Sign in to answer this question. Sign in to comment. Even if this plane and line is not intersecting, it shows check=1 and intersection point I =[-21.2205 31.6268 6.3689]. 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 … There are three possibilities: The line could intersect the plane in a point. Do you mean lines or line segments? Example \(\PageIndex{9}\): Other relationships between a line and a plane, \[\begin{align*} \text{Line:}\quad x &=1 + 2t & \text{Plane:} \quad x + 2y - 2z = 5 \\[5pt] y &= -2 + 3t \\[5pt] z &= -1 + 4t \end{align*}\nonumber\]. So the point of intersection can be determined by plugging this value in for \(t\) in the parametric equations of the line. 0 ⋮ Vote. If we take the parameter at being one of the coordinates, this usually simplifies the algebra. 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. 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. Distinguishing these cases and finding the intersection point have use, for example, in computer graphics, motion planning, and collision detection. This is equivalent to the conditions that all . A given line and a given plane may or may not intersect. For and , this means that all ratios have the value a, or that for all i. Determine whether the following line intersects with the given plane. 0 Comments . Finally the code uses the adjusted values of t1 and t2 to find those closest points. Or the line could completely lie inside the plane. What if we keep the same line, but modify the plane equation to be \( x + 2y - 2z = -1\)? Here: \(x = 2 - (-3) = 5,\quad y = 1 + (-3) = -2, \,\text{and}\quad z = 3(-3) = -9\). Here you can calculate the intersection of a line and a plane (if it exists). 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. How can we tell if a line is contained in the plane? If the line does intersect with the plane, it's possible that the line is completely contained in the plane as well. In Euclidean geometry, the intersection of a line and a line can be the empty set, a point, or a line. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), [ "article:topic", "authorname:pseeburger", "license:ccby" ], \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\). 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). Do a line and a plane always intersect? For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Missed the LibreFest? 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. The code then adjusts t1 and t2 so they are between 0 and 1. But the line could also be parallel to the plane. The point of intersection is a common point of a line and a plane. Follow 40 views (last 30 days) Stephanie Ciobanu on 9 Nov 2017. Can you please explain what is the issue? Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Have questions or comments? 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 \( x + 2y - 2z = -1\). 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 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)? 0. Intersection of plane and line. Therefore, coordinates of intersection must satisfy both equations, of the line and the plane. Substituting the expressions of \(t\) given in the parametric equations of the line into the plane equation gives us: \[(1+2t) +2(-2+3t) - 2(-1 + 4t) = 5\nonumber\]. Example \(\PageIndex{8}\): Finding the intersection of a Line and a plane. This means that this line does not intersect with this plane and there will be no point of intersection. 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. We can verify this by putting the coordinates of this point into the plane equation and checking to see that it is satisfied. Line: x = 2 − t Plane: 3 x − 2 y + z = 10 y = 1 + t z = 3 t. If we take the parameter at being one of the coordinates, this usually simplifies the algebra. No. You say "lines" but you say they have length. If t1 and t2 are both between 0 and 1, then the line segments intersect. Watch the recordings here on Youtube! 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 … Can i see some examples? In 2D, with and , this is the perp prod… Legal. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Those values generate the points on the two segments that are closest to the point of intersection. If they do intersect, determine whether the line is contained in the plane or intersects it in a single point. Check: \(3(5) - 2(-2) + (-9) = 15 + 4 - 9 = 10\quad\checkmark\). 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 \(t = -3\). Or the line could completely lie inside the plane. 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. 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. The parametric equation of a line, x = x0 + at, y = y0 + bt and z = z0 + ct

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