Question

write the equation of a line and a plane that intersect at the point (1,4,-1)?
pleassseee help the textbook gives an answer of x+2y+5z-4=0

Answer 1

You can pick any line, any plane. Let's say I want my line's direction to be parallel to the vector \((1,1,0)\). Then we can write a vector-valued function that describes the line: \(\mathbf{r}(t)=(1,4,-1)+t(1,1,0)\quad\implies x=1+t,y=4+t,z=-1\).

Now, let's say I want my plane orthogonal to the line. This means the direction vector will be our plane's normal. Recall that any vector lying in the plane must be orthogonal to the normal, and any point in the plane yields a vector from \((1,4,-1)\) pointing towards it:$$\mathbf{x}=(x,y,z)\\\mathbf{x}_0=(1,4,-1)\\(\mathbf{x}-\mathbf{x}_0)\cdot(1,1,0)=0\\(x-1)+(y-4)=0\\x+y-5=0$$
Said plane and line intersect at one and only one point.$$1+t+4+t-5=0\\2t=0\\t=0$$... which corresponds to our point \(\mathbf{r}(0)=(1,4,-1)\).

Answer 2

Well said @oldrin.bataku !

Answer 3

thank you!!!