Let the direction vector of the line be u = (p, q, r). To find u and a second point on the line, compare the x - X1 equation of the line 2 y - 4 = z with the standard form of the equation y - Y1 z - z1 -1 p r (X1, Y1, Z4) ,4, 0 2 -1, 1) u = Step 2 Let v be the vector from the point (0, 4, 0) to the point (12, -10, 6), then v can be given in the compon form as (8,0, – 16) × v = (12, – 14, 6) Step 3 You have found u = (2, -1, 1) and v = (12, -14, 6). Find the normal vector for the planen = u x v. i 1 2 n = u x v = -1 12 -14 V -14 = 8i - 16 16 k Step 4 The plane containing the point (x2, Y2, Z2) = (12, -10, 6) and having the normal vector (8, 0, –16) = (a, b, c) can be represented in the standard form as, n = 8i – 16k = a(x — х2) + b(y — У2) + c(z — z2) 3 о. = Substitute in the equation of the plane and simplify. 8(х — ) + (y + 10) – 16(z – 6) = 0