Cálculo

CHAPTER 12

Three-Dimensional Space; Vectors
EXERCISE SET 12.1
1. (a) (0, 0, 0), (3, 0, 0), (3, 5, 0), (0, 5, 0), (0, 0, 4), (3, 0, 4), (3, 5, 4), (0, 5, 4) (b) (0, 1, 0), (4, 1, 0), (4, 6, 0), (0, 6, 0), (0, 1, −2), (4, 1, −2), (4, 6, −2), (0, 6, −2) 2. corners: (2, 2, ±2), (2, −2, ±2), (−2, 2, ±2), (−2, −2, ±2) z (–2, –2, 2) (–2, 2, 2) (–6, 1, –2) (2, –2, 2) (–2, –2, –2) (2, –2, –2) (2, 2, 2) y y (4, 1, 1) x (2, 2, –2) (4, 1, –2) x (4, 2, 1)

3. corners: (4, 2, −2), (4,2,1), (4,1,1), (4, 1, −2), (−6, 1, 1), (−6, 2, 1), (−6, 2, −2), (−6, 1, −2) z (–6, 2, 1) (–6, 2, –2)

(–2, 2, –2)

4. (a) (x2 , y1 , z1 ), (x2 , y2 , z1 ), (x1 , y2 , z1 )(x1 , y1 , z2 ), (x2 , y1 , z2 ), (x1 , y2 , z2 ) (b) The midpoint of the diagonal has coordinates which are the coordinates of the midpoints 1 of the edges. The midpoint of the edge (x1 , y1 , z1 ) and (x2 , y1 , z1 ) is (x1 + x2 ), y1 , z1 ; 2 the midpoint of the edge (x2 , y1 , z1 ) and (x2 , y2 , z1 ) is of the edge (x2 , y2 , z1 ) and (x2 , y2 , z2 )) is midpoint of the diagonal are 5. (a) a single point on that line (b) a line in that plane (c) a plane in 3−space 6. (a) R(1, 4, 0) and Q lie on the same vertical line, and so does the side of the triangle which connects them. R(1, 4, 0) and P lie in the plane z = 0. Clearly the two sides are perpendicular, and the sum of the squares of the two sides is |RQ|2 + |RP |2 √ 42 + (22 + 32 ) = 29, = so the distance from P to Q is 29. z Q (1, 4, 4)

1 x2 , (y1 + y2 ), z1 ; the midpoint 2

1 x2 , y2 , (z1 + z2 ) . Thus the coordinates of the 2

1 1 1 (x1 + x2 ), (y1 + y2 ), (z1 + z2 ) . 2 2 2

y R(1, 4, 0) P (3, 1, 0) x

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Exercise Set 12.1

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(b) S(3, 4, 0) and P lie in the plane z = 0, and so does SP . S(3, 4, 0) and Q lie in the plane y = 4, and so does SQ. Hence the two sides |SP | and |SQ| are perpendicular, and |P Q| = |P S|2 + |QS|2 = 32 + (22 + 42 ) = 29.

z Q (1, 4, 4)

y

P (3, 1, 0) x

S(3, 4, 0)

(c) T (1, 1, 4) and Q lie on a