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To find the intersection of line and a plane.

Line: $x=y-1=5z$ and Plane : $4x-y+3z=17$.

Let $x=y-1=5z =t$, then $x=t, y=t+1$ and $z = t/5$. Thus puttinf this into the equation of plane we have $ t =5.$

Thus the line intersects the plane at $(5, 6,1)$.

Is the solution correct?

User8976
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1 Answers1

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Your solution is correct.

Vector $(1, 1, \frac15)$ is parallel to the line: $x=y-1=5z$ & $(4, -1,3)$ is normal to the plane: $4x-y+3z=17$.

The dot product: $(1, 1, \frac15)\cdot (4, -1,3)=\frac{18}{5}\ne 0$ shows that the line: $x=y-1=5z$ intersects the plane at an angle ($\ne 90^\circ$ i.e. not parallel to the plane)

Now, plug in $x=t, y=t+1, z=t/5$ into equation of plane: $4x-y+3z=17$

$$4(t)-(t+1)+3(\frac{t}{5})=17$$ $$\frac{18t}{5}=18$$ $$t=5$$ The line intersects the plane at a single point $(5, 6,1)$

Harish Chandra Rajpoot
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