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Equation of a Plane Passing through Three Points

Find the vect...

Question

Find the vector equation of the plane passing through the point $^i +^j - 2^k,^i + 2^j +^k, 2^i -^j +^k$ . Hence find the cartesian equation of the plane

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Solution

$\to a, \to b, \to c$ are the position vectors

i.e. $\to a =^i +^j - 2^k, \to b =^i + 2^j +^k, \to c = 2^i -^j +^k$

and $\to b - \to a = 0^i +^j + 3^k$

$\to c - \to a =^i - 2^j + 3^k$

Now, $(\to b - \to a) \times (\to c - \to a) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 0 & 1 & 3 1 & - 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$=^i (3 + 6) -^j (0 - 3) +^k (0 - 1)$

$= 9^i + 3^j -^k$

Let $\to n = (\to b - \to a) \times (\to c - \to a) = 9^i + 3^j -^k$

Then, the vector equation of the required plane is $\to r \cdot \to n = \to a \cdot \to n$

$\to r \cdot (9^i + 3^j -^k) = (^i +^j - 2^k) \cdot (9^i + 3^j -^k)$

$\to r \cdot (9^i + 3^j -^k) = 9 + 3 + 2$

$\to r \cdot (9^i + 3^j -^k) = 14$

The cartesian equation of the plane is given by,

$(x^i + y^j + z^k) \cdot (9^i + 3^j -^k) = 14 ∵ [\to r = x^i + y^j + z^k]$

$9 x + 3 y - z = 14$

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