Find the equation of the plane through Po(-0.5, 1, 1) with normal vector n - [8] 3
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- Determine the equation for the line with normal vector n = [3,-4] passing through the point P(-1,-2)Find a vector n normal to the plane with the equation 6z – (5x + 5y) = −1. n = (-5, >Calculate the distance from point B=(−1,3,−1) to the plane that passes through point A=(3,2,−1) and has normal direction n =(5,4,4).
- Find an equation of the plane through (1, −3, 5) with normal vector n =〈2, 1, −4〉.Find an equation of the plane that passes through the point P(-3, -2, 3) and has the vector n = (-7, 6, 4) as a normal.Find both the parametric and vector equation of the line segment between (−2, 3) and (5, −1) where 0 ≤ t ≤ 5. Please explain your steps. Thank you.
- Write the scalar equation of the line given the normal vector n =[1,−1] and a point Po (-5,1)Determine the equation of the tangent plane and a vector equation of the normal line to x²y - 4ze+y +35= 0 at (3,-3,2). A. tangent plane: -26x + y −4z +89 = 0, normal line: (3+26t, -3-t, 2+4t) B. tangent plane: -26x + y - 4z +89 = 0, normal line: (3-26t, -3+t, 2-4t) C. tangent plane: 26x + y − 4z +89 = 0, normal line: (3+26t, -3-t, 2+4t) D. tangent plane: 26x + y - 4z +89 = 0, normal line: (3-26t, -3+t, 2-4t)Find the linear equation of the plane through the point (2, −2, 6) and with normal vector j+2k.
- Find the value of d in the equation of the plane ax+by+cz+d=0 passing through the point P(3, 1, -2) and has the vector n=(-3, -5, -9) as a normal. %3DFind the equation of the plane in xyz-space through the point P = (3, 5, 3) and perpendicular to the vector n = = (−1, −3, 3). 2 =Find the equation of a plane that passes has a normal vector through (-1,4,1) and that is parallel to the line x=4-t₁y = 4t, 2=1+t.