# Physics 221 Calc-based question for homework

• AidenPhysica
In summary, Ray Vickson says that the result of taking the cross product of two vectors is always zero if the vectors are perpendicular to each other.
AidenPhysica

## Homework Statement

If vector d1--> is equal to 2i^-4j^+9k^ and vector d2-> is equal to 8i^-10j^+9k^, what is (d1-->+d2-->) * (d1--> X 4d2->)?

## The Attempt at a Solution

So I added up first two vectors and then multiplied it by the cross product but I get a vector and when I find its length its like 1931. I just want to know if this is even close.

Hi AidenPhysica, Welcome to Physics Forums.

Your result does not look correct to me. Can you show the details of your calculations?

$$(\vec{d_1} + \vec{d_2}) \cdot (\vec{d_1} \times 4 \vec{d_2})$$
?

If that is correct, then your result is wrong. What to you get when you take the cross product of two vectors?

Its right what you put except there has to be a 4 right in front of d2 so its multiplied by four.

AidenPhysica said:
Its right what you put except there has to be a 4 right in front of d2 so its multiplied by four.
I've corrected it. So, what do you get when you take the cross product of two vectors?

Ok, so this is what I did. I first added up the two vectors to get 10i+-14j+18k. Then you do the cross product, so I did that operation of the cross product, which is basically how much two vectors are perpendicular to each other? So here is the matrix i j k on top second row is 2, -4, 9. Third row is 32i-40j+36k. The cross product of this matrix is

216i +216j+48k

I'll give you a hint: the actual values of the vectors are not important.

AidenPhysica said:
216i +216j+48k
What is the orientation of that vector with respect to the original vectors?

Then what I did was I multiplied each respective component of the first answer times the second to get 3815 after doing pythagrean theorem is that better?

OK, so what do you mean by orientation?

AidenPhysica said:
Then what I did was I multiplied each respective component of the first answer times the second to get 3815 after doing pythagrean theorem is that better?
Stop a minute. I want you to think about the problem first. You can calculate numerical values later.

AidenPhysica said:
OK, so what do you mean by orientation?
The direction in which it points. When you take the cross product of two vectors, the resulting vector points in a very specific direction with respect to the two other vectors.

so isn't the cross product equal to magnitude of a times magnitude of b multiplied by sin of theta?

Isn't cross product equal to magnitude of a times magnitude of b times sin of theta?

so isn't it just: ok so we have 10.05 times 26.83 times sin of theta=309.22. So then theta=309.22/269.6*(arcsin)

I don't know it doesn't work.

I try plugging it into a times b sin theta = magnitude of cross product but it just doesn't work.

so how do i find orientation though?

Is it in degrees? I am just confused I see the picture, its like perpendicular.

AidenPhysica said:
I am just confused I see the picture, its like perpendicular.
Yes, the resulting vector is perpendicular to the plane of the two initial vectors.

I'm sorry can you please tell me at least the next step, i am confused and have been at this for hours.

AidenPhysica said:
Then what I did was I multiplied each respective component of the first answer times the second to get 3815 after doing pythagrean theorem is that better?

(1) What is the general formula for ##\mathbf{c}= \mathbf{a} \times \mathbf{b}, ## where ##\mathbf{a} = a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k}## and ##\mathbf{b} = b_x \mathbf{i} + b_y \mathbf{j} + b_z \mathbf{k}?## Express your answer as ##\mathbf{c} = c_x \mathbf{i} + c_y \mathbf{j} + c_z \mathbf{k},## and give explicit formulas for ##c_x, c_y, c_z## in terms of the ##a##'s and ##b##'s. If you do not know the formula, you can Google "vector cross product" to find hundreds of articles that will be helpful.
(2) What is the geometric interpetation of ##\mathbf{c}## in relation to ##\mathbf{a}## and ##\mathbf{b}?##
(3) How can you do the question without doing any calculations at all?

Thank you all for the help, thank you Ray Vickson and the mentors as well. So Mr. Vickson, I found the answer to be 0 via calculations but I only vaguely understand why it is 0. Is it because whenever you take dot product, which is a scalar of how much one vector lies along another vector, and take the dot product within the cross product of those same vectors, which is the vector perpendicular to the two vectors, you cancel the two values out and it just leaves you with 0?

AidenPhysica said:
Thank you all for the help, thank you Ray Vickson and the mentors as well. So Mr. Vickson, I found the answer to be 0 via calculations but I only vaguely understand why it is 0. Is it because whenever you take dot product, which is a scalar of how much one vector lies along another vector, and take the dot product within the cross product of those same vectors, which is the vector perpendicular to the two vectors, you cancel the two values out and it just leaves you with 0?

Use the elementary fact that
$$(\mathbf{d}_1 + \mathbf{d}_2) \cdot \mathbf{c} = \mathbf{d}_1 \cdot \mathbf{c} +\mathbf{d}_2 \cdot \mathbf{c}$$
Here, ##\mathbf{c} = \mathbf{d}_1 \times 4 \mathbf{d}_2 =4 ( \mathbf{d}_1 \times \mathbf{d}_2 ).##

Can you see why ##\mathbf{d}_1 \cdot \mathbf{c}## and ##\mathbf{d}_2 \cdot \mathbf{c}## are both equal to 0? (Hint: question number (2) in my response.)

Last edited:

## What is Physics 221 Calculus-Based?

Physics 221 Calculus-Based is a course that combines the principles of physics with the mathematical concepts of calculus. It is designed to provide students with a deeper understanding of how physical systems work and how to solve complex problems using calculus.

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The topics covered in Physics 221 Calculus-Based typically include kinematics, Newton's laws of motion, work and energy, linear and rotational motion, oscillations and waves, and basic principles of thermodynamics and electricity.

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