How do you find a unit vector with positive first coordinate that is orthogonal to the plane through the points $P = (3, - 3, 0), Q = (5, - 1, 2)$ $P = (3, -3, 0), Q = (5, -1, 2)$ , and $R = (5, - 1, 6)$ $R = (5, -1, 6)$ ?

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How do you find a unit vector with positive first coordinate that is orthogonal to the plane through the points $P = (3, - 3, 0), Q = (5, - 1, 2)$ $P = (3, -3, 0), Q = (5, -1, 2)$ , and $R = (5, - 1, 6)$ $R = (5, -1, 6)$ ?

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Answer 1 · 2016-09-08T10:09:01

$\frac{\sqrt{2}}{2} {1, - 1, 0}$ $sqrt(2)/2{1,-1,0}$

Explanation:

If $P = (3, - 3, 0), Q = (5, - 1, 2)$ $P = (3, -3, 0), Q = (5, -1, 2)$ , and $R = (5, - 1, 6)$ $R = (5, -1, 6)$ pertain to the plane $Pi$

then $(P-Q) xx (R-Q)$ is normal to $Pi$

so

$vec n = (P-Q) xx (R-Q) = {-8, 8, 0}$ . Any vector proportional to $vec n$ like $lambda vec n$ with $lambda in RR, lambda ne 0$ is also normal to $Pi$ so we choose

$lambda = -1/norm(vec n) = -1/(8 sqrt(2))$ . The sought unit vector is then

$lambda vec n =sqrt(2)/2{1,-1,0}$

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How do you find a unit vector with positive first coordinate that is orthogonal to the plane through the points $P = (3, - 3, 0), Q = (5, - 1, 2)$ $P = (3, -3, 0), Q = (5, -1, 2)$ , and $R = (5, - 1, 6)$ $R = (5, -1, 6)$ ?

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Explanation:

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How do you find a unit vector with positive first coordinate that is orthogonal to the plane through the points P = (3, -3, 0), Q = (5, -1, 2)P=(3,−3,0),Q=(5,−1,2)P = (3, -3, 0), Q = (5, -1, 2), and R = (5, -1, 6)R=(5,−1,6)R = (5, -1, 6)?

1 Answer

Explanation:

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How do you find a unit vector with positive first coordinate that is orthogonal to the plane through the points $P = (3, - 3, 0), Q = (5, - 1, 2)$ $P = (3, -3, 0), Q = (5, -1, 2)$ , and $R = (5, - 1, 6)$ $R = (5, -1, 6)$ ?