{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions5

# Solutions5 - MATH 214 A1 Assignment 5 Due Friday June 10...

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

MATH 214 A1 - Assignment 5 Due Friday, June 10, 5:00pm (1) (a) Find the equation of the plane containing the points (0 , 0 , 2), (1 , 0 , 0) and (1 , - 1 , 1). (2 marks) 2 x + y + 2 (b) Find the distance between the plane calculated in part ( a ) and the point (3 , 0 , 10). (3 marks) Let P 0 = (0 , 0 , 2) and P = (3 , 0 , 10), then d = P 0 P · n | n | = h 3 , 0 , 8 i · h 2 , 1 , 1 i |h 2 , 1 , 1 i = 14 6 (2) Given the vector-valued function r ( t ) = h 4 9 (1+ t ) 3 2 , 4 9 (1 - t ) 3 2 , 1 3 t i : (a) Find r 0 ( t ) (2 marks) r 0 ( t ) = h 2 3 1 + t, - 2 3 1 - t, 1 3 i (b) Find a ( t ) (2 marks) a ( t ) = h 1 3 1 + t , 1 3 1 - t , 0 i (c) Find the angle between the two vectors. (3 marks) cos( θ ) = r · a | r || a | = 0 | r || a | = 0 Thus the two vectors are perpendicular. (3) If an object starts at rest with initial position vector h 0 , 0 , 0 i at t = 1 and undergoes acceleration described by a ( t ) = h 1 t , 1 5 - t , 1 2 t i from t = 1 to t = 3: (a) Find the velocity at time t = 2 (2 marks) The velocity vector is given by R a ( t ) dt , or v ( t ) = h ln t, - ln | 5 - t | , 1 2 ln t i + C using the condition that v (1) = h 0 , 0 , 0 i , we find that C = h 0 , ln 4 , 0 i so v ( t ) = h ln t, - ln | 5 - t | + ln 4 , 1 2 ln t i thus the velocity at time 2 is v (2) = h ln 2 , ln (

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 / 3

Solutions5 - MATH 214 A1 Assignment 5 Due Friday June 10...

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

View Full Document
Ask a homework question - tutors are online