# Thread:

1. ## Finding unit vectors

Directions: Find a unit vector in the direction of line PQ.
P(7, -4) Q(-3, 2)

the textbook say the answer is: 1/SquareRoot(34) * (-5, 3)
sorry I don't know how to make the square root symbol...

ok, so I know how to find unit vectors for single points. can anyone please explain how to do this one? is it similar to the other method?

Thanks!

2. find the vector PQ first which is (-10 , 6) , then divide by the magnitude of the vector to get a unit vector.

3. Originally Posted by bobak
find the vector PQ first which is (-10 , 6) , then divide by the magnitude of the vector to get a unit vector.

This might be a stupid question, but how did you get (-10, 6) ??

4. To find a unit vector in any direction, one divides by the length in that direction.
If V is a non-zero vector then $\frac{1}{{\left\| V \right\|}}<br />$ is a unit vector parallel to V.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9272693395614624, "perplexity_flag": "head"}