Find the equation of the plane passing through the line of intersection of the planes and and parallel to x-axis.

Open in App

Solution

The given planes are

The equation of any plane passing through the line of intersection of these planes is

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.

Substituting in equation (1), we obtain

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

Suggest Corrections

Similar questions

\to r \cdot (2^i - 3^j + 4^k) = 1

\to r \cdot (^i -^j) + 4 = 0

\to r \cdot (2^i -^j -^k) + 4 = 0,

\to r \cdot (^i + 3^j -^k) = 9

\to r \cdot (2^i -^j +^k) = 3

\to r \cdot (^i +^j +^k) = 6

\to r \cdot (2^i + 3^j + 4^k) = - 5

(1, 1, 1)

Find the equation of the plane which contains the line of intersection of the planes , and which is perpendicular to the plane .

(1, 2, 3)

\to r \cdot (^i -^j + 2^k) = 5

\to r \cdot (3^i +^j +^k) = 6

Join BYJU'S Learning Program