Find the vector equation of the plane containing the point A(1, -2, 7) and the z-axis.
Answer: r = (1, -2, 7) + s(0, 0, 1) + t(1, -2, 6)
i don't get which direction vectors you use... I'm getting confused. Can someone help me?
51 minutes
Answer: r = (1, -2, 7) + s(0, 0, 1) + t(1, -2, 6)
i don't get which direction vectors you use... I'm getting confused. Can someone help me?
51 minutes
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Given the position vector xₒ of a point P and two directions d1 and d2, the vector equation of the plane through P containing d1 and d2 is r = xₒ + sd1 +td2, where s and t are scalars.
This comes from completing the triangle of vectors formed by P and any point r in the desired plane. The sides of the triangle are || to and scalar multiples of d1 and d2.
We can take P as (1,−2,7) and d1 = (0,0,1). For d2, take any point Q on the z-axis ( say (0,0,1) ) and let d2 = PQ = (1,−2, 6). This gives desired equation.
This is a bit contrived to give stated answer. Undirected, I would have taken origin (0,0,0) as P and d2 = PO = (1,−2,7) so plane would be r = u(0, 0, 1) + v(1, −2, 7). Putting u = s−t and v = t+1 gives quoted result but this is a simpler equation.
This comes from completing the triangle of vectors formed by P and any point r in the desired plane. The sides of the triangle are || to and scalar multiples of d1 and d2.
We can take P as (1,−2,7) and d1 = (0,0,1). For d2, take any point Q on the z-axis ( say (0,0,1) ) and let d2 = PQ = (1,−2, 6). This gives desired equation.
This is a bit contrived to give stated answer. Undirected, I would have taken origin (0,0,0) as P and d2 = PO = (1,−2,7) so plane would be r = u(0, 0, 1) + v(1, −2, 7). Putting u = s−t and v = t+1 gives quoted result but this is a simpler equation.