Prove (by using a point and a normal vector) that the plane whose intercepts with the coordinate axes are x=1, y=1, and z=1 is given by the equation x+y+z=1
To prove that the plane with intercepts x=1, y=1, and z=1 is given by the equation x+y+z=1, we can use the point-normal form of the equation of a plane.
Let's consider a point P on the plane, and let's choose the point P(1, 1, 1). We want to show that this point lies on the plane.
Now, let's find a normal vector to the plane. Since the plane intercepts with the coordinate axes are x=1, y=1, and z=1, we can take the direction ratios of the normal vector as the reciprocal of these intercepts. Therefore, the normal vector N = (1/1, 1/1, 1/1) = (1, 1, 1).
The equation of a plane in point-normal form is given by:
N · (Q - P) = 0,
where N is the normal vector, Q is any point on the plane, and P is a known point on the plane.
Plugging in the values for N and P, we get:
(1, 1, 1) · (Q - (1, 1, 1)) = 0.
Find the dot product:
(Q1 - 1) + (Q2 - 1) + (Q3 - 1) = 0.
Simplifying this expression, we get:
Q1 + Q2 + Q3 = 1.
Hence, we have shown that any point Q that satisfies the equation Q1 + Q2 + Q3 = 1 lies on the plane with intercepts x=1, y=1, and z=1.
Therefore, the plane with the intercepts x=1, y=1, and z=1 is given by the equation x+y+z=1.