Math Solution 1

# Math Solution 1 - Homework due 26th January Solutions to...

This preview shows pages 1–2. Sign up to view the full content.

Homework due 26th January Solutions to compulsory questions Problem ( § 1.3, Exercise 24). Find an equation for the plane that passes through the point (2 , - 1 , 3) and is perpendicular to the line v = (1 , - 2 , 2) + t (3 , - 2 , 4). Solution. The plane has a normal vector 3 i - 2 j + 4 k , and so has equation given by ( ( x i + y j + z k ) - (2 i - j + 3 k ) ) · (3 i - 2 j + 4 k ) = 0 . Equivalently, 3 x - 2 y + 4 z = 20 . Problem ( § 1.3 Exercise 32). Given vectors a and b , do the equations x × a = b and x · a = k a k determine a unique vector x ? Argue geometrically and analytically. There are quite a few ways to approach this problem. Here are a few. Solution 1. Firstly, if a = 0 then any vector x satisﬁes the equations, and so the solution is not unique in this case. Suppose now that a 6 = 0 . If there are two solutions, x and y say, then x · a = k a k y · a = k a k and x × a = b y × a = b . Subtracting gives ( x - y ) · a = 0 and ( x - y ) × a = 0 , and so x - y is both perpendicular and parallel to a . Therefore x - y = 0 , and so x = y . Therefore the solution, if it exists, is unique provided a 6 = 0 . (The question did not actually ask you to prove the existence of a solution). Solution 2. A more geometric argument proceeds by determining the length and direction of x . If

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

Math Solution 1 - Homework due 26th January Solutions to...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online