Combinations of Lines & Planes (Edexcel A Level Further Maths: Core Pure): Revision Note

Intersections of lines & planes

How do I tell if a line is parallel to a plane?

• A line is parallel to a plane if its direction vector is perpendicular to the plane’s normal vector

• If you know the Cartesian equation of the plane in the form $ax+by+cz=d$ then the values of a, b, and c are the individual components of a normal vector to the plane

• The scalar product can be used to check in the direction vector and the normal vector are perpendicular

  • If two vectors are perpendicular their scalar product will be zero

How do I tell if the line lies in the plane?

• If the line is parallel to the plane then it will either never intersect or it will lie in the plane

  • Check to see if they have a common point

• If a line is parallel to a plane and they share any point, then the line lies in the plane

How do I find the point of intersection of a line and a plane in Cartesian form?

• If a line is not parallel to a plane it will intersect it at a single point

• If both the vector equation of the line and the Cartesian equation of the plane is known then this can be found by:

• STEP 1: Set the position vector of the point you are looking for to have the individual components x, y, and z and substitute into the vector equation of the line

  • $\left(\genfrac{}{}{0pt}{}{x}{\begin{array}{c}y\\ z\end{array}}\right)= \left(\genfrac{}{}{0pt}{}{x_0}{\begin{array}{c}{y}_0\\ z_0\end{array}}\right)+\lambda \left(\genfrac{}{}{0pt}{}{l}{\begin{array}{c}m\\ n\end{array}}\right)$

• STEP 2: Find the parametric equations in terms of x, y, and z

  • $x= x_0+ \lambda l$

  • $y= y_0+ \lambda m$

  • $z= z_0+ \lambda n$

• STEP 3: Substitute these parametric equations into the Cartesian equation of the plane and solve to find λ

  • $a\left(x_0+ \lambda l\right)+b(y_0+ \lambda m)+c(z_0+ \lambda n)=d$

• STEP 4: Substitute this value of λ back into the vector equation of the line and use it to find the position vector of the point of intersection

• STEP 5: Check this value in the Cartesian equation of the plane to make sure you have the correct answer

How do I find the point of intersection of a line and a plane in vector form?

• Suppose you have a line with equation $\left(\genfrac{}{}{0pt}{}{x}{\begin{array}{c}y\\ z\end{array}}\right)= \left(\genfrac{}{}{0pt}{}{x_0}{\begin{array}{c}{y}_0\\ z_0\end{array}}\right)+t\left(\genfrac{}{}{0pt}{}{l}{\begin{array}{c}m\\ n\end{array}}\right)$ and plane with equation $\left(\genfrac{}{}{0pt}{}{x}{\begin{array}{c}y\\ z\end{array}}\right)= \left(\genfrac{}{}{0pt}{}{a_1}{\begin{array}{c}{a}_2\\ a_3\end{array}}\right)+\lambda \left(\genfrac{}{}{0pt}{}{b_1}{\begin{array}{c}{b}_2\\ b_3\end{array}}\right)+\mu \left(\genfrac{}{}{0pt}{}{c_1}{\begin{array}{c}{c}_2\\ c_3\end{array}}\right)$

• Form three equations with unknowns t, λ and μ

• Solve them simultaneously on your calculator

• Substitute the values back in to get the intersection

Angle between a line & a plane

How do I find the angle between a line and a plane?

• When you find the angle between a line and a plane you will be finding the angle between the line itself and the line on the plane that creates the smallest angle with it

  • This means the line on the plane directly under the line as it joins the plane

• It is easiest to think of these two lines making a right-triangle with the normal vector to the plane

  • The line joining the plane will be the hypotenuse

  • The line on the plane will be adjacent to the angle

  • The normal will be opposite to the angle

• As you do not know the angle of the line on the plane you can instead find the angle between the normal and the hypotenuse

  • This is the angle opposite the angle you want to find

  • This angle can be found because you will know the direction vector of the line joining the plane and the normal vector to the plane

  • This angle is also equal to the angle made by the line at the point it joins the plane and the normal vector at this point

• The smallest angle between the line and the plane will be 90° minus the angle between the normal vector and the line

  • In radians this will be $\frac{\mathrm{\pi }}{2}$ minus the angle between the normal vector and the line