How would you find a unit vector with positive first coordinate that is orthogonal to the plane through the points P = (3, -4, 4), Q = (6, -1, 7), and R = (6, -1, 9)?

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Answer 1 · 2017-10-04T01:10:15

We need to create two vectors in the plane.

To create $vec(PQ)$ , we subtract each coordinate of point P from its respective coordinate of point Q:

$vec(PQ) = < 6-3, -1 - (-4), 7 - 4>$

$vec(PQ) = < 3, 3, 3>$

To create $vec(QR)$ , we subtract each coordinate of point P from its respective coordinate of point R:

$vec(PR) = < 6-3, -1 - (-4), 9-4 >$

$vec(PR) = < 3, 3, 5 >$

A normal vector to the plane, $vecn$ , is the cross product of these two vectors:

$vecn = vec(PQ) xx vec(PR)$

$vecn = < 3, 3, 3 > xx < 3, 3, 5>$

$vecn = <6, -6, 0>$

To make $vecn$ a unit vector, we divide by the magnitude:

$|vecn| = sqrt(6^2+ (-6)^2+0^2)$

$|vecn| = 6sqrt2$

$hatn = 1/(6sqrt2)<6, -6, 0>$

$hatn = sqrt2/(12)<6, -6, 0>$

$hatn = < sqrt2/2, -sqrt2/2, 0 >$