Given 3 points given by position vectors a - b -k- c -k- Find the plane passing through these 3 points-A- r-i-3 -k-1B- r-i-3 -k-1C- r-i-3 -k-2D- r-i-3 -k-2

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Equation of Plane Passing through Three Points

Given 3 point...

Question

Given 3 points given by position vectors $¯ a =^i +^j, ¯ b =^j +^k, ¯ c = -^i -^j -^k$ . Find the plane passing through these 3 points.

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Solution

The correct options are
A
B
First we have to find the vector perpendicular to the plane. It will be along $(¯ a - ¯ b) \times (¯ b - ¯ c)$
$(¯ a - ¯ b) \times (¯ b - ¯ c) = ((^i +^j) - (^j +^k) \times (^j +^k) - (-^i -^j -^k)) = (^i -^k) \times (^i + 2^j + 2^k) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 0 & - 1 1 & 2 & 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i (0 + 2) -^j (2 + 1) +^k (2 - 0) = 2^i - 3^j + 2^k \Rightarrow^n o r d i r e c t i o n p e r p e n d i c u l a r t o t h e p l a n e n e e d e d i s \frac{2^i - 3^j + 2^k}{\sqrt{17}}$
Without loss of generality we will choose $¯ a$ to be the point on the plane. Therefore equation of the plane will be $(¯ r - ¯ a) .^n = 0$
$(¯ r - (^i +^j)) .^n = 0 (¯ r - (^i +^j)) . \frac{(2^i - 3^j + 2^k)}{\sqrt{17}} = 0 ¯ r . (¯ ¯¯¯ ¯ 2 i - 3^j + 2^k) = (^i +^j) . (2^i - 3^j + 2^k) ¯ r . (2^i - 3^j + 2^k) = 2 - 3 = - 1 ∴ ¯ r . (¯ i - 3^j +^k) = - 1$
This can also be written as $¯ r . (- ¯ i + 3^j -^k) = 1$
Therefore both a and b are correct.

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Given 3 points given by position vectors ¯a=^i+^j,¯b=^j+^k,¯c=−^i−^j−^k. Find the plane passing through these 3 points.

Given 3 points given by position vectors $¯ a =^i +^j, ¯ b =^j +^k, ¯ c = -^i -^j -^k$ . Find the plane passing through these 3 points.