Consider the plane defined by equation $3x+4y-z=2$ and a line defined by the following vector equation (in parametric form)

$x(t)=2-2t$, $y(t)=-1+3t$, $z(t)=-t$

(a) Find the point where the line intersects the plane.

(b) Find the normal vector to the plane.

(c) Find the angle at which the line intersects the plane (Hint: Use dot product)

My workings:

(a) I have found the point of intersection at $(2,-1,0)$ by substituting the parametric vector equation into the equation of the plane.

(b) Normal is $(3,4,-1)$

(c) I'm a little stumped here. Do I use this formula $a.b=|a||b|\cos\theta$ to solve for the angle?

  • $\begingroup$ First two is correct. For part $(c)$, yes you use that identity for dot product. $\endgroup$
    – MathNewbie
    Commented May 28, 2015 at 6:04
  • $\begingroup$ But somehow I could not get the answer given (π/2) - arccos ((√91)/26) @MathNewbie $\endgroup$
    – Larry Lee
    Commented May 28, 2015 at 6:06

1 Answer 1


The normal to the plane is ${\bf n}=(3,4,-1)$ as you have found. A vector in the direction of the line is ${\bf v}=(-2,3,-1)$. The angle between them is given by the dot product formula: $$\cos\theta=\frac{\bf n\cdot v}{|{\bf n}|\,|{\bf v}|}=\frac{7}{\sqrt{26}\sqrt{14}} =\frac{7}{2\sqrt{91}}=\frac{\sqrt{91}}{26}\ .$$ And the angle you want is $\frac\pi2-\theta$, draw a diagram and you will see why.


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