Anonymous From: - Posts: - Votes: - |
1.1 Write down an equation of the planewhich passes through the point (-9,-3,5) that isperpendicular to the vector [1,-3,5]. Leave your answer in the form A x + B y + C z = D, where A, B, Care integers. Answer: 1.2 Find an equation of the plane whichpasses through the point (7,-8,3) thatis parallel to the plane -6 x-5 y+8 z = 0. Leave your answer in the form A x + B y + C z = D, where A, B, Care integers. Answer: |

TinySupa4 From: US Posts: 14 Votes: 28 |
1.1 x-3y+5z-25=0 : point - normal form or point distance formof a plane. A=1, B=-3, C=5 and D=25. 1.2 -6x-5y+8z=22. |

Anonymous From: - Posts: - Votes: - |
1.1 the plane is perpendicular to the vector[1,-3, 5]. So the vector is the normalof the plane, we have 1*x - 3*y + 5*z = D, or x - 3y +5z = D the plane passes through the point(-9,-3, 5),so (-9) - 3*(-3) + 5*5 = D = 25 answer: x - 3y + 5z = 25 1.2 the plane is parallel to the plane -6 x-5 y+8 z = 0. so it can be written as -6x - 5y + 8z = D it passes through the point (7,-8, 3), so -6*7 - 5*(-8) + 8*3 = D = 22 answer: -6x - 5y + 8z = 22 |

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