How to Find a Unit Vector Perpendicular to Another Vector 8i + 4j - 6k

In this video, we talk about how to find a unit vector perpendicular to another vector. We are going to present two ways to do this; one using dot product and the other using cross product.
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[@dalarev9858]

Thanks for the second method. It's useful to avoid dividing by 0.
Which drawing program is that? It's neat that drawn lines are automatically straightened.

[@GlassofNumbers]

I used an app on Samsung laptop called Noteshelf. Thank you for the comment!!

[@JKMath]

Congrats on 2k subscribers!

[@GlassofNumbers]

Thank you! I am working toward 3k now 😁

[@user-mm6vm2kc6p]

I prefer second method and thanks I like your video is helpful 👏

[@GlassofNumbers]

Thank you! Please help share this video with others 😁👍

[@osinachiaugustine4635]

Thanks for the approach one

[@GlassofNumbers]

Thank you for your comment! Please share my video with others 😁

[@Mahmudul5610]

I like your teaching

[@GlassofNumbers]

Thank you! Please help me share the video with others 👍😁

[@Jordannahuwo622]

For approach number 2, how do we work out a vector in 2 dimensions?

[@GlassofNumbers]

If a vector lies in the xy-plane, for example, v = (1, 3), then we can choose z = 0. That is, (1, 3, 0) in the space.

[@Jordannahuwo622]

Thanks that makes sense

[@daoodveer5853]

thanks sir g

[@GlassofNumbers]

You are welcome! Definitely help me share the videos with others 😁

[@douglaspetrus312]

Why is it that you choose 1,1,2 in the first approach

[@GlassofNumbers]

I chose that because the numbers are easy. We can choose other numbers since there are infinitely many vectors perpendicular to .

[@shivam_646]

Isn't there a way to solve it properly without picking any random no.

[@GlassofNumbers]

Both approaches are proper ways to solve this problem. In fact, there is no problem with picking random numbers since the answer is not unique for this question.

[@Wize.]

cant abc just equal 0?

[@GlassofNumbers]

But the zero vector is not a unit vector.

[@Wize.]

@@GlassofNumbers oh okok thanks for the response🙏