$\begingroup$

Two numbers $3$ and $4$ their multiplication is each one from the first number is repeated a number of times as the second number i.e. $3$ times $4$ is $(1+1+1)$ times four meaning $1+1+1+1 + 1+1+1+1 + 1+1+1+1$ for example we can represent $3$ by any countable thing like three bags each of them has four apples then the total is $12$ .. Excuse me for this I know it's obvious but I want an explanation like this one I've given

Assuming we have two one dimensional vectors $ 3\hat{i}$ and $3 \hat{j}$ then Could you give me explain it to me in the previous way I explained multiplication of scalars ..

$\endgroup$ 18

2 Answers

$\begingroup$

Let's turn this around. If you wanted to know (for general vectors) what vector is perpendicular to both of them, you would have to solve two equations with three unknowns. You know that the dot product of two vectors is zero if they are perpendicular, so you would write

$$\vec{x}\cdot \vec{a} = 0\\ \vec{x}\cdot \vec{b} = 0$$

But if you are working in three dimensions, that leaves you with two equations and three unknowns - so you can't solve it.

We could add a third constraint. For example, we could decide that the length of the third vector has to be equal to the area of the parallelogram described by the first two vectors. We know the area is $A=|\vec{a}|\cdot|\vec{b}|\cdot \sin\theta$. That gives me the length of vector $\vec{x}$ and my third equation.

Now I have three equations and three unknowns. I could write them out in full- the first two are easy:

$$x_1 a_1 + x_2 a_2 + x_3 a_3 = 0\\ x_1 b_1 + x_2 b_2 + x_3 b_3 = 0$$

The third one is harder. Following , if you set the length of vector $\vec{x}$ equal to 1, you could solve the equations but they are quite messy:

$$\vec{x} = ±\frac{1}{Z}(a_3b_2 - a_2 b_3, a_1b_3 - a_3 b_1, a_2 b_1 - a_1 b_2)$$

Then some more manipulation tells you that if you set the length equal to the magnitude of the area instead, the expression simplifies to the one we know and love.

Feel free to read through the lines and lines of derivation at the link above... it is long because it assumes very little advanced algebra - but that seems to be what you are asking for.

$\endgroup$ 1 $\begingroup$

Find me a vector that's mutually orthogonal to two other vectors

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy