Question

The equation of the plane passing through the point (-1,-2,-3) and perpendicular to the x-axis is

• A. x=-1
• B. y=-2
• C. z=-3
• D. 2x+3y=5
• E. x+y+z=6

Answer 1

The Correct Option is A

The equation of a plane is generally given by the formula: \[ Ax + By + Cz + D = 0 \] where \( A, B, C \) are the direction ratios of the normal to the plane. In this case, the plane is perpendicular to the x-axis. A plane perpendicular to the x-axis has a normal vector along the x-axis, so the direction ratios are \( A = 1, B = 0, C = 0 \). Thus, the equation of the plane is: \[ x = -1 \] This plane passes through the point \( (-1, -2, -3) \), and since it is parallel to the yz-plane, the equation of the plane is simply \( x = -1 \).

The correct option is (A) : \(x=-1\)

Answer 2

A plane perpendicular to the x-axis has a normal vector parallel to the x-axis. The direction vector of the x-axis is \(\hat{i} = (1, 0, 0)\).

Therefore, the normal vector to the plane is \(\bar{n} = (1, 0, 0)\).

The equation of a plane with normal vector \(\bar{n} = (A, B, C)\) passing through a point \((x_0, y_0, z_0)\) is given by \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\).

In our case, \(\bar{n} = (1, 0, 0)\) and \((x_0, y_0, z_0) = (-1, -2, -3)\). So the equation of the plane is:

\(1(x - (-1)) + 0(y - (-2)) + 0(z - (-3)) = 0\)

\(1(x + 1) + 0 + 0 = 0\)

\(x + 1 = 0\)

\(x = -1\)

Therefore, the equation of the plane is \(x = -1\).