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Exercise 6.7 · Q6

Q.Find the parametric vector, non-parametric vector and Cartesian form of the equations of the plane passing through the three non-collinear points (3,6,−2),(−1,−2,6)(3,6,-2),(-1,-2,6), and (6,4,−2)(6,4,-2).

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Take one point as base, form the two edge vectors to the other two points, cross them for the normal, then write all three requested forms of the plane's equation.

Step 1. Base point and edge vectors. A(3,6,−2),B(−1,−2,6),C(6,4,−2)A(3,6,-2),B(-1,-2,6),C(6,4,-2): b⃗−a⃗=(−4,−8,8)\vec b-\vec a=(-4,-8,8), c⃗−a⃗=(3,−2,0)\vec c-\vec a=(3,-2,0).

Step 2. Parametric vector equation.

r⃗=(3i^+6j^−2k^)+s(−4i^−8j^+8k^)+t(3i^−2j^).\vec r=(3\hat i+6\hat j-2\hat k)+s(-4\hat i-8\hat j+8\hat k)+t(3\hat i-2\hat j).

Step 3. Normal =(b⃗−a⃗)×(c⃗−a⃗)=(\vec b-\vec a)\times(\vec c-\vec a).

∣i^j^k^−4−883−20∣=i^(0+16)−j^(0−24)+k^(8+24)=16i^+24j^+32k^.\begin{vmatrix}\hat i&\hat j&\hat k\\-4&-8&8\\3&-2&0\end{vmatrix}=\hat i(0+16)-\hat j(0-24)+\hat k(8+24)=16\hat i+24\hat j+32\hat k.

Simplify (divide by 88): normal ∝(2,3,4)\propto(2,3,4). …

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