Find the equation of plane passing through points $(1, 1, - 1) (6, 4, - 5) (- 4, - 2, 3)$ .

Open in App

Solution

Vector equation of a plane passing through three points with position vectors $\to a, \to b, \to c$ is
$\begin{matrix} (\to r - \to a) \cdot [(\to b - \to a) \times (\to c - \to a)] = 0 N o w, t h e p l a n e p a s s e s t h r o u g h t h e p o i n t s A (1, 1, - 1) \to a = 1 ˆ i + 1 ˆ j - 1 ˆ k B (6, 4, - 5) \to b = 6 ˆ i + 4 ˆ j - 5 ˆ k C (- 4, - 2, 3) \to c = - 4 ˆ i - 2 ˆ j + 3 ˆ k (\to b - \to a) = (6 ˆ i + 4 ˆ j - 5 ˆ k) - (1 ˆ i + 1 ˆ j - 1 ˆ k) = (6 - 1) ˆ i + (4 - 1) ˆ j + (- 5 - (- 1)) ˆ k = 5 ˆ i + 3 ˆ j - 4 ˆ k (\to c - \to a) = (- 4 ˆ i - 2 ˆ j + 3 ˆ k) - (1 ˆ i + 1 ˆ j - 1 ˆ k) = (- 4 - 1) ˆ i + (- 2 - 1) ˆ j + (3 - (- 1)) ˆ k = - 5 ˆ i - 3 ˆ j + 4 ˆ k (\to b - \to a) \times (\to c - \to a) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 5 & 3 & - 4 - 5 & - 3 & 4 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - ∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 5 & 3 & - 4 5 & 3 & - 4 \end{matrix} ∣ ∣ ∣ ∣ ∣ = \to 0 \end{matrix}$
This implies, the three points are collinear.
vector equation of plane is
$[\to r - (ˆ i + ˆ j - ˆ k)] \cdot \to 0 = 0$
Since, the above equation is satisfied for all values of $\to r$
Therefore, there will be infinite planes passing through he given collinear points.

Suggest Corrections

Similar questions

(1, 1, - 1), (6, 4, - 5), (- 4, - 2, 3)

(4, - 3)

(1, 1)

(2, 3)

Join BYJU'S Learning Program