How To Find Where A Line Intersects A Plane

2. Intersection points of a line with $xy$ airplane

Intersection points of a line with $xy$ aeroplane

The plan is:

(1) Observe out the parametric equations of the 2 known lines;

(2) The unknown line intersects the ii known lines. This yields two atmospheric condition for the parametric equation of the unknown line. Find out these two weather condition.

(3) The unknown line passes through the point (0,0,1). This reveals in part the equation of the unknown line.

(4) The unknown line intersects the xy-plane (or, the airplane $z=0$). Using this condition, utilize the to a higher place derived equation of the unknown line to make up one's mind the x and y-coordinate of the intersection point. Its z-coordinate is $z=0$.

In item,

Line (1) is given equally an intersection of two planes,

$$x+2y+z=1$$ $$-x+y-2z=2$$

That is, the first plane has a normal $\vec{n_1}$= (one,ii,1), while the second plane has a normal $\vec{n_2}$= (-one,1,-2).

Then the vector $\vec{v}=\vec{n_1}\times\vec{n_2}$=(-5,-i,3) is parallel to line (1), and its parametric equation is

$$\vec{ten}=\vec{x_0}+\vec{v}t$$

We can see that $\vec{v}$ is not parallel to any of the coordinate planes thus it is safe to assume that line (one) intersects the aeroplane $y=0$. Plug this into the equation of the ii intersecting planes defining line (i) and observe the intersection point to be (4,0,-3). Therefore, the parametric equation of line (1) is:

$$x=4-5t$$ $$y=-t$$ $$z=-three+3t$$

Analogously, process the known equations of the ii intersecting planes, defining line (2): The vector $\vec{v}$ in this case is (1,-ane,-one). Line (2) intersects the coordinate plane $y=0$ in betoken (2,0,0) which gives the parametric equation of line(two) to exist:

$$x=2+s$$ $$y=-due south$$ $$z=-s$$

Now, assume the unknown line has parametric equation

$$x=x_0+pa$$ $$y=y_0+atomic number 82$$ $$z=z_0+pc$$

This line intersects Line (1): at the intersection indicate, $x$ and $y$ are identical. Therefore:

$$four-5t=x_0+pa$$ $$-t=y_0+pb$$

From here, limited $t$, and $p$ through the unknowns $x_0,y_0,a,b$. At present, $z$ must be identical for both lines at the intersection bespeak, or else the lines are skew. In the equations for $z$, now plug in $t$ and $p$ to notice the get-go status:

$$z_0+c\frac{x_0-5y_0-4}{5b-a}=-3-3y_0+3b\frac{x_0-5y_0-iv}{a-5b}$$

Analogously, Line (two) and the unknown line intersect and using the aforementioned procedure, are establish to yield at the intersection point the 2d condition, namely

$$z_0+c\frac{2-x_0-y_0}{a+b}=y_0-b\frac{x_0+y_0-2}{a+b}$$

Because the unknown line passes through the point (0,0,one), we tin can write its parametric equation every bit

$$ten=pa$$ $$y=pb$$ $$z=1+pc$$

where $x_0=0, y_0=0,z_0=i$. Plug these into conditions 1 and two in a higher place. Express $c$. Plug its value into the commencement condition and solve for $b$:

$$b=\frac{a}{three}.$$

Then, $c=-\frac{a}{3}$. Finally, the parametric eq. of the unknown line is

$$x=pa$$ $$y=\frac{pa}{three}$$ $$z=i-\frac{pa}{3}$$

Use these equations to find out, that this line intersects the coordinate airplane $z=0$ at point (3,1,0).

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Comments

• Question

  A line passing through $(0,0,one)$ and intersecting lines $x+2y+z=1 , -x+y-2z =2$ and $x+y =2 ,x+z=2$. What is the intersecting points of the line with $xy$ aeroplane

  My effort

  I computed the line of intersection of both the pair of plane

  $$L_1= -5i + j + 3k$$

  $$L_2 = i - j - grand$$

  So tried to become the direction ratios (let it exist $ai+bj+ck$) of the line parallel to the line nosotros need to find via the status that all 3 lines need to be coplanar to intersect. Which gives me

  $$a-b+2c=0$$

  And I am stuck hither

  • I have added and answer below. I hope this helps. Please re check all expressions

Recents

Source: https://9to5science.com/intersection-points-of-a-line-with-xy-plane

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