Question

is dot product of a vector with itself equals to modulus of that vector?

Answer 1

@thomaster @sleepyjess @Compassionate

Answer 2

I assume modulus = length in the text you're using.

As a quick way to answer this yourself: What is the length of < 1, 1, 1 > ? And how did you compute it?

Answer 3

\[\langle x_1,x_2\rangle \cdot \langle x_1,x_2\rangle = x_1^2 + x_2^2 = \|\langle x_1,x_2\rangle \|^2 \]

Answer 4

@mathteacher1729
\[\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}\]=modulus of vector

and dot product of <1,1,1> is also \[\sqrt{3}\]
so is this right

Answer 5

Not quite. The dot product is of <1,1,1> with itself is:

\[
\langle 1,1,1 \rangle \cdot \langle 1,1,1 \rangle = 1+1+1 = 3
\]

So the dot product is not the length of the vector. The dot product is the *square of the* length of the vector, as @rational pointed out in the general case. :)

Answer 6

ok i got it with the help of you both @rational @mathteacher1729
thanks to clear my doubt