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Find a point-normal form equation of the plane containing (1, -1, 4) and line with parametric equations x=1+3t, y=2+6t, and z=−t
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- Provide the point-normal form equation of the plane containing (1, -1, 4) and line with parametric equations x=1+3t, y=2+6t, and z=−tFind the distance from the point (3, 2, 5) to the plane whose general equation is 2x + 3y - z = 0.Provide the parametric equations of the normal line to the surface defined by the vector equation given at the point (0,0,π).
- Determine the equation of the plane that contains the point (1, 1, 1) and intersects the yz plane along the line z = 1 - y.What is the equation of the plane passing through the point (2, -3, 1) and perpendicular to the line with parametric equations x = 1 + t, y = 2 - 2t, z = 3t?Find parametric equations for the normal line to the surface z=-7x2 -9y2 at the point (2,1,-37)
- Find the general equation of the plane through the point (3, 2, 5) that is parallel to the plane whose general equation is 2x + 3y -z = 0.Give the general equation of the plane satisfying the given condition. plane Σ containing the point (5, 0, −7) and parallel to the plane given by2x − 6y + 3z = 0Find an equation of the plane containing the point (2,1,-1) andparallel to the plane 3x – y + 2z = 1
- Find an equation of the plane in standard form (ax + by + cz = d) that passes through the point (4, 0, −2) and contains the line given parametrically by x = 2 − 7t, y = 6 + 5t, z = 4 + 8t.Find the set of symmetric and parametric equations of the line passing through the point P(3, 4, 1) that is parallel to the plane 3x−2y −z −4 = 0 and perpendicular to the line 2x = −y/2 = −3z.Locate the point of intersection of the plane 2x + y - z = 0 and the line through (4,2,0) that is perpendicular to the plane. If it is not possible to locate such points, explain why?