Question

Find the distance between the point and the line/plane 

a. (-1,4);x-3y+2=0

b. (-1,-1,2);2x+5y-6z=4

Answer 1

Distance between point (x₁, y₁) and the line ax + by + c = 0 is (ax₁ + by₁ + c)/$\sqrt{(a^2+b^2)}$

Distance between point (–1, 4) and the line x – 3y + 2 = 0 is |(–1 – 3 × 4 + 2)/$\sqrt{(1^2+(-3)^2)}$ |

                                                                                            = |–11/$\sqrt{10}$ |

                                                                                            = 3.4785

 

 

Distance between point (x₁, y₁, z₁) and the plane ax + by + cz + d = 0 is (ax₁ + by₁ + cz₁)/$\sqrt{(a^2+b^2+c^2)}$ )

Distance between point (–1, –1, 2) and the plane 2x + 5y –6z – 4 = 0 is |(2(–1) + 5(–1) – 6 × 2)/$\sqrt{(2^2+5^2+(-6)^2)}$ |

                                                                                       = |–19/$\sqrt{65}$ |

                                                                                       = 2.3567