点对于某个对称轴镜像翻转，这里用二维演示，思路及外围办法都是统一的，二维能较好阐明。

找到在对称轴上与点最近的点

 // 将对称轴整成线段
 const line = new THREE.Line3(axis, new THREE.Vector3(0, 0, 0))
 const linePoint = new THREE.Vector3()
 line.closestPointToPoint(arr[i], false, linePoint)
 // linePOint 就是 arr[i] 点在线段上最近的点 

得出点与线段上的点的间隔 S1

// linePoint 是线段上的点  vec3 是须要翻转的点
const distance = linePoint.distanceTo(vec3)

按方向缩短 S1 的间隔至 S 点

 // 简略的勾股定理
 const p = Math.sqrt(2) * 0.5
 x = linePoint.x + p * distance
 y = linePoint.y - p * distance

须要进行判断

对于上方的后果只是点在对称轴左侧才成立, 须要判断点的方位

  /**
   * 判断是否在左侧
   * 因为数据是否在左右侧关系到 xy 的符号问题, 间隔是没有正数的，而坐标是存在正负的
   */
  function leftORRight() {
    let left = true;
    const line = new THREE.Line3(axis, new THREE.Vector3(0, 0, 0))
    const linePoint = new THREE.Vector3()
    for (let i = 0; i < arr.length; i++) {line.closestPointToPoint(arr[i], false, linePoint)
      if (linePoint.x < arr[i].x) {
        // 右侧
        left = false;
        break;
      }
    }
    return left;
  }

All Code

import * as THREE from 'three'
/**
 * 实现点绕制订对称轴镜像转换的性能
 */
function axisRound(scene) {
  // const arr = [//   new THREE.Vector3(-1, 1, 0),
  //   new THREE.Vector3(-2, -1, 0),
  //   new THREE.Vector3(-1, -1, 0),
  //   new THREE.Vector3(-1, 2, 0),
  // ]
  // 提供测试的数据
  const arr = [new THREE.Vector3(1, -1, 0),
    new THREE.Vector3(2, -1, 0),
    new THREE.Vector3(1, 1, 0),
    new THREE.Vector3(1, -2, 0),
  ]
  const material = new THREE.MeshBasicMaterial({color: 'red', side: THREE.DoubleSide});
  
  arr.forEach(row => {const geometry = new THREE.PlaneGeometry(0.5, 0.5);
    const plane = new THREE.Mesh(geometry, material);
    scene.add(plane);
    plane.position.set(row.x, row.y, row.z)
  })
  const axis = new THREE.Vector3(1, 1, 0)
  {const points = [];
    points.push(axis);
    points.push(new THREE.Vector3(-10, -10, 0))
    points.push(new THREE.Vector3(10, 10, 0))
    const geometry = new THREE.BufferGeometry().setFromPoints(points);
    const line = new THREE.Line(geometry, material);
    scene.add(line);
  }
  arr.forEach(row => {const geometry = new THREE.PlaneGeometry(0.5, 0.5);
    const plane = new THREE.Mesh(geometry, material);
    scene.add(plane);
    const n = t(row, axis)
    console.log(n);
    plane.position.set(n.x, n.y, n.z)
  })
  // 正式开始计算
  function t(vec3, axis) {const line = new THREE.Line3(axis, new THREE.Vector3(0, 0, 0))
    const linePoint = new THREE.Vector3()
    line.closestPointToPoint(vec3, false, linePoint)
    // 得出 2 的算术平方根的 1 /2
    const p = Math.sqrt(2) * 0.5
    const distance = linePoint.distanceTo(vec3)
    let x, y;
    if (leftORRight()) {
      x = linePoint.x + p * distance
      y = linePoint.y - p * distance
    } else {
      x = linePoint.x - p * distance
      y = linePoint.y + p * distance
    }
    return new THREE.Vector3(x, y, 0)
  }
  /**
   * 判断是否在左侧
   * 因为数据是否在左右侧关系到 xy 的符号问题, 间隔是没有正数的，而坐标是存在正负的
   */
  function leftORRight() {
    let left = true;
    const line = new THREE.Line3(axis, new THREE.Vector3(0, 0, 0))
    const linePoint = new THREE.Vector3()
    for (let i = 0; i < arr.length; i++) {line.closestPointToPoint(arr[i], false, linePoint)
      if (linePoint.x < arr[i].x) {
        // 右侧
        left = false;
        break;
      }
    }
    return left;
  }
}
export {axisRound}