Question

Find an equation of the plane passing through (0 comma negative 1 comma 1 )that is orthogonal to the planes 3xplus4ynegative 3zequals0 and negative 3xplus2yplus4zequals5.

Answer 1

Answer:

The equation of plane is

$22x -3y+18z = 21$

Step-by-step explanation:

We have to find the equation of plane passing through the point (0,-1,1) and orthogonal to the planes

$n_1:3x + 4y-3z = 0\\n_2:-3x + 2y + 4z = 5$

Thus, we can write:

$n_1:\\n_2:$

We will evaluate:

$n = n_1\times n_2\\\\n = \left[\begin{array}{ccc}i&j&k\\3&4&-3\\-3&2&4\end{array}\right] \\\\n = i(16 + 6)-j(12-9) +k(6+12)\\n = 22i-3j+18k\\n =$

The required plane passes through the point (0,-1,1)

Thus, the equation of plane is

$n.=0\\\Rightarrow .=0\\\Rightarrow 22(x) -3(y+1) + 18(z-1) =0\\\Rightarrow 22x - 3y - 3 + 18z - 18 = 0\\\Rightarrow 22x -3y+18z = 21$

is the required equation of the plane.