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Find the line that passes through the point $\left(2,5,3\right)$ and is perpendicular to the plane $2x-3y+4z+7=0$

Workings:

The normal vector of the plane is $\vec{n} = \langle2,-3,4\rangle$

The equation of the line in vector form is given by: $$\vec{r} = (2,5,3) + t\langle2,-3,4\rangle$$

Confusion:

The dot product of the line and direction of the plane is not equal to zero:

$$\begin{pmatrix} 2 \ -3 \ 4 \end{pmatrix} \cdot \begin{pmatrix} 2 \ -3 \ 4 \end{pmatrix} = 4 + 9 + 16 = 29 \neq 0$$

Does this mean that the line is not perpendicular to the plane? According this question, if the direction of the line and normal of plane are scalar multiples of each other, then the line is perpendicular.

-https://math.stackexchange.com/questions/741488/determine-whether-the-line-and-plane-are-perpendicular

Since the normal of the plane and direction of the line are equal, can a vector be considered a scalar multiple of itself?

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2 Answers 2

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The dot product of the line and direction of the plane is not equal to zero

This is true, in fact if $x$ is a vector, by the properties of the dot product you have $x\cdot x=0 \iff x=0$. This does not mean that the line and the plane are not perpendicular, in fact the right condition is:

If the direction of the line and normal of plane are scalar multiples of each other, then the line is perpendicular.

Now, coming to your question:

Since the normal of the plane and direction of the line are equal, can a vector be considered a scalar multiple of itself?

Sure, this is the simplest case and the scalar multiple is $1$.

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Let $A$ & $B$ be 2 Arbitrary Points on that Plane. The line between these two Points must be Perpendicular to Normal that you have got.

Hence , we can take $A$ with $y=0$,$z=0$ , to get $2x−0+0+7=0$ hence $x=-7/2$.
We can take $B$ with $x=0$,$z=0$ , to get $0−3y+0+7=0$ hence $y=7/3$.

Vector $\vec{AB}$ is $(-7/2,0,0)-(0,7/3,0)=(-7/2,-7/3,0)$

Check : Is that vector Perpendicular to what you got ?
$(2,-3,4)\cdot(-7/2,-7/3,0)=-7+7+0=0$

We have the Verification that you have the Correct Normal.
We can Double verify with new Point C $x=0$,$y=0$ , to get $0−0+4z+7=0$ hence $z=-7/4$.
You can then try Dot-Product $\vec{BC}$ with Normal to get 0.
Cross-Product $\vec{CA}$ with Normal must be 0. Every line on the Plane will have $0$ Dot-Product with the Normal.

"Confusion:
Dot Product of the line and Direction of the Plane is not equal to 0"

Yes : It must not be 0. It will almost never be 0.
That Line & the Direction of Plane [[Normal]] are like Parallel , not Perpendicular.

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