Linear Algebra Solutions 88

# Linear Algebra Solutions 88 - Hence if C =(1 0 1 then the...

This preview shows page 1. Sign up to view the full content.

Hence if C = (1 , 0 , 1), then the closest point on N to C is given by P = A + t - AB , where t = ( - AC · - AB ) /AB 2 . Now - AC = 3 / 2 - 3 / 2 1 and - AB = 5 / 2 - 1 / 2 1 , so t = 3 2 × 5 2 + - 3 2 × - 1 2 + 1 × 1 ( 5 2 ) 2 + ( - 1 2 ) 2 + 1 2 = 11 15 . Hence P = - 1 / 2 3 / 2 0 + 11 15 5 / 2 - 1 / 2 1 = 4 / 3 17 / 15 11 / 15 , so P = (4 / 3 , 17 / 15 , 11 / 15). Also the shortest distance PC is given by PC = s 1 - 4 3 2 + 0 - 17 15 2 + 1 - 11 15 2 = 330 15 . 9. The intersection of the planes x + y - 2 z = 4 and 3 x - 2 y + z = 1 is the line given by the equations x = 9 5 + 3 5 z, y = 11 5 + 7 5 z, where z is arbitrary. Hence the line L has a direction vector [3 / 5 , 7 / 5 , 1] t or the simpler [3 , 7 , 5] t . Then any plane of the form 3 x + 7 y + 5 z = d will be perpendicualr to
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern