MA1E02 Tutorial sheet [January 18 – 22, 2016] Name: Student ID:

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MA1E02 Tutorial sheet

[January 18 – 22, 2016]

Name: Student ID:

1.

(a) Show (on the graph) the point P with co ordinates (2 , 4 , 1) and the point ordinates Q (1 , 2 , with co 5) [and sketch in boxes to show where the points are in space].

(b) Calculate the distance from P to Q .

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4 5 3 4 Solution: By the distance formula p ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 = = p (1 − 2) 2 + (2 √ 1 + 4 + 16 = − 4) 2 √ 21 + (5 − 1) 2 is the distance from P to Q .

2. Find the equation of the points ( x, y, z ) that are on the sphere of radius 2 and centre (3 , − 4 , 5) . Find an answer without square roots. [Hint: the points on the sphere are those with distance from the center exactly equal to the radius.] Solution: If we write the description of the points on the sphere as a formula, we get distance (( x, y, z ) , (3 , − 4 , 5)) = 2 p ( x − 3) 2 + ( y + 4) 2 + ( z − 5) 2 = 2 ( x − 3) 2 + ( y + 4) 2 + ( z − 5) 2 = 4

We have squared both sides to eliminate the square root.

t

Note

: Normally we could introduce extra solutions to an equation by squaring both sides.

The equation t = 4 has just the one solution, but t 2 = 16 has two solutions: t = 4 and = − 4 . As the distance between points is never negative, we can find the distance if we know its square. So we do not introduce any extra solutions by squaring both sides.

3. Find the centre and radius of the shere with equation x 2 + y 2 + z 2 − 2 x + 6 y − z = 0 [Hint: Complete squares like x 2 − 2 x = ( x − 1) 2 − 1 .] Solution: x 2 − 2 x y 2 + 6 y z 2 − z x 2 + y 2 + z 2 − 2 x + 6 y − z = ( x − 1) 2 − 1 = ( y + 3) 2 − 9 = ( z − 1 / 2) 2 − 1 / 4 = ( x − 1) 2 + ( y + 3) 2 + ( z − 1 / 2) 2 − 41 / 4 So the equation x 2 + y 2 + z 2 − 2 x + 6 y − z = 0 can be rewritten ( x − 1) 2 + ( y + 3) 2 + ( z − 1 / 2) 2 = 41 4 or or Hence centre (1 , − 3 , 1 / 2) ( dist (( x, y, z ) , (1 , − 3 , 1 / 2))) 2 dist (( x, y, z ) , (1 , − 3 , 1 / 2)) = 41 = 4 √ 41 2 and radius √ 41 / 2 .

Richard M. Timoney 2

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