{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

s11wk03su - Math 23 B Dodson Week 2 Homework 12.5 Lines...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 23 B. Dodson Week 2 Homework: 12.5 Lines, Planes 12.6 Quadratic Surfaces Problem 12.5.3: Give vector and (scalar) parametric equations for the line through the point (-2,4,10) parallel to the vector < 3 , 1 , 8 > . Solution: The vector equation vector OP = vector OP 0 + t vector d, for when the position vector of the point P ( x,y,z ) puts P on the line through P 0 ( x 0 ,y 0 ,z 0 ) with direction vector vector d = < a,b,c > gives < x,y,z > = < x 0 ,y 0 ,z 0 > + t < a,b,c > = < - 2 , 4 , 10 > + t < 3 , 1 , 8 >, which we can view as a “point-slope” equation, where P 0 is the point, and vector d gives the direction of the line. (Here in the position vector vector OP, O = O (0 , 0 , 0) is the Origin.) To get the scalar equations, we use scalar mult. and vector add to write the vector equation
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 . as < x,y,z > = < - 2 + 3 t, 4 + t, 10 + 8 t >, and simply read-off x = - 2 + 3 t, y = 4 + t, z = 10 + 8 t. Here, each value of the parameter t gives a point on the line. Problem 12.5.26: Find an equation of the plane through the point ( - 2 , 8 , 10) and perpendicular to the line x = 1 + t, y = 2 t, z = 4 - 3 t.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern