Question

Find an equation of the plane. the plane that passes through the point (−1, 1, 2) and contains the line of intersection of the planes x + y − z = 2 and 4x − y + 5z = 3

Answer 1

To find an equation of the plane, first determine the direction ratios of the line of intersection of the given planes. Then, substitute a point on the plane into the equation to find the constant term.

Step-by-step explanation:

To find an equation of the plane, we need to determine its normal vector and a point that lies on the plane. The line of intersection of the planes x + y − z = 2 and 4x − y + 5z = 3 will lie in the plane we are looking for.

First, find the direction ratios of the line of intersection by equating the two planes:

• x + y − z = 2
• 4x − y + 5z = 3

Solving these equations, we get the direction ratios of the line of intersection as (13, 21, 5). So, the direction ratios of the line of intersection are also the direction ratios of the plane.

Next, we substitute the point (−1, 1, 2) into the equation of the plane to find the constant term. Using the direction ratios we found earlier, the equation of the plane is 13(x+1) + 21(y-1) + 5(z-2) = 0.

Answer 2

Step-by-step explanation:

The plane containing line of intersection of two planes x+y-z-2=0 and 4x-y+5z-3 =0

will be of the form

x+y-z-2+k(4x-y+5z-3) =0

Now given that the plane passes through (-1,1,2)

Substitute to get

-1+1-2-2 +k(-4-1+10-3) =0

-4+2k =0

Or k =2

Substitute in the equation of the plane for k to get real equation.

x+y-z-2+2(4x-y+5z-3) =0

Simplify to get

9x-y+9z-8 =0 is the equation of the required plane.