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I'm reading from Marsden Vector Calculus 6th Edition and this picture is from page 43.

enter image description here

I'm having difficulty understanding how they get to $$ \text{Distance} =|\vec v \cdot \vec n|$$

The way I tried to work it out is using the projection formula:

$$\operatorname{proj}_{\vec n} \vec v= \frac{\vec v \cdot \vec n}{(\frac{A\vec i + B\vec j + C\vec k}{\sqrt{A^2+B^2+C^2}})^2} \cdot \frac{A\vec i + B\vec j + C\vec k}{\sqrt{A^2+B^2+C^2}} $$

$$\operatorname{proj}_{\vec n} \vec v= \frac{\vec v \cdot \vec n}{\frac{A\vec i + B\vec j + C\vec k}{\sqrt{A^2+B^2+C^2}}}$$

$$\operatorname{proj}_{\vec n} \vec v= \frac{\vec v \cdot \vec n}{\vec n}$$

and since $\vec n $ is a unit vector, we know it equals to $1$ so $$\operatorname{proj}_{\vec n} \vec v= \vec v \cdot \vec n$$

Question 1:But I still don't get where they got the absolute values from around the distance formula.

Question 2:Also hoe does the projection of $\vec v$ onto $\vec n$ make sense when projections are defined as dropping a perpendicular from the end of vector $\vec b$ toa line parallel to vector $\vec a$ in case of $proj_{\vec a} \vec b$. However in the case in the derivation we have that the line dropped from $\vec v$ is parallel(not perpendicular) to $\vec n$. Why is that?

To give an idea of what I mean by the last question:enter image description here The second picture was taken from: http://tutorial.math.lamar.edu/Classes/CalcII/DotProduct.aspx

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  • $\begingroup$ You have $\frac{\vec{v}\cdot \vec{n}}{\vec{n}}$ which makes no sense. You can't divide by a vector! Division by a vector simply isn't defined. Perhaps you mean $|\vec{n}|$ in the denominator. That is NOT an "absolute value", it is the length of the vector $\vec{n}$. $\endgroup$
    – user247327
    Feb 1, 2019 at 22:53

1 Answer 1

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Question 1: The dot product can be negative. If the point E is "below" the plane with respect to the given orientation, then without the absolute value, the answer would be negative. The absolute value makes it positive.

Question 2: The projection of a vector $b$ onto a vector $a$ $proj_a b$ is parallel to $a$. The "line dropped" is perpendicular to $a$ and, when subtracted from $b$, yields $proj_a b$ which is then parallel to $a$. In the picture shown, the solid black line is the "line dropped from $v$," and the dotted black line is the projection of $v$ onto the span of $n$.

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