Find the equation of the plane passing through the point (1,-1,2) having 2,3,2 as direction ratios of normal to the plane.

Open in App

Solution

The plane passes through the point having position vetor $\to a = ˆ i - ˆ j + 2 ˆ k$ and is normal to the vector $\to n = 2 ˆ i + 3 ˆ j + 2 ˆ k$ . So, the vector equation of the plane is
$(\to r - \to a) . \to n = 0$
$\to r . \to n = \to a . \to n$
$\to r . (2 ˆ i + 3 ˆ j + 2 ˆ k) = (ˆ i - ˆ j + 2 ˆ k) . (2 ˆ i + 3 ˆ j + 2 ˆ k)$
$\to r . (2 ˆ i + 3 ˆ j + 2 ˆ k) = 3$
The cartesian equation of the plane is $2 x + 3 y + 2 z = 3$ .

Suggest Corrections

Similar questions

(- 1, - 2, 1)

(1, 3, - 2)

(x_{1}, y_{1}, z_{1})

a, b, c

(0, 1, 1), (1, 1, 2)

(- 1, 2, - 2)

(0, 1, 1), (1, 1, 2)

(- 1, 2, - 2)

Join BYJU'S Learning Program