Question

Find the vector equation of the plane through the points (2, 1, - 1) and (- 1, 3,4) and perpendicular to the plane x - 2y + 4z = 10.

Answer 1

The required plane passes through two points P(2, 1, - 1) and Q(- 1, 3, 4).

Let \(\overrightarrow a and \overrightarrow b\)  be the position vectors of points P and Q, respectively.

Then, \(\overrightarrow a\) = 2î + ĵ - k̂

and \(\overrightarrow b\) = - î + 3ĵ + 4k̂

Now, \(\overrightarrow {PQ} = \overrightarrow b - \overrightarrow a \) = (- î + 3ĵ + 4k̂) - (2î + ĵ - k̂)

= - 3î + 2ĵ + 5k̂

Let \(\overrightarrow {n_1}\) be the normal vector to the required plane.

Then,

\(\overrightarrow {n} = \overrightarrow {n_1} \times \overrightarrow {PQ} =\) \(\begin{vmatrix} \hat i & \hat j & \hat k \\[0.3em] 1 & -2 & 4 \\[0.3em] -3 & 2 & 5 \end{vmatrix}\)

= î(- 10 - 8) - ĵ(5 + 12) + k̂(2 - 6)

= - 18î - 17ĵ - 4k̂

The required plane passes through a point having position vector \(\overrightarrow a\) = 2î + ĵ - k̂ and normal vector \(\overrightarrow n\) = - 18î - 17ĵ - 4k̂ . So, its vector equation is

 \((\overrightarrow r - \overrightarrow a) . \overrightarrow n \implies \overrightarrow r .\overrightarrow n = \overrightarrow a . \overrightarrow n\)

⇒ \(\overrightarrow r\). (- 18î - 17ĵ - 4k̂) = (2î + ĵ - k̂) (- 18î - 17ĵ - 4k̂)

⇒ \(\overrightarrow r\). (- 18î - 17ĵ - 4k̂) = - 36 - 17 + 4

∴ \(\overrightarrow r\). (18î + 17ĵ + 4k̂) = 49