What is the unit vector that is orthogonal to the plane containing $(3i + 2j - 3k)$ and $(i -2j + 3k)$ ?

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Answer 1 · 2016-12-19T07:25:35

The answer is $=〈0,-3/sqrt13,-2/sqrt13〉$

We do a cross product to find the vector orthogonal to the plane

The vector is given by the determinant

$| (hati,hatj,hatk), (3,2,-3), (1,-2,3) |$

$=hati(6-6)-hatj(9--3)+hatk(-6-2)$

$=〈0,-12,-8〉$

Verification by doing the dot product

$〈0,-12,-8〉.〈3,2,-3〉=0-24+24=0$

$〈0,-12,-8〉.〈1,-2,3〉=0+24-24=0$

The vector is orthgonal to the other 2 vectors

The unit vector is obtained by dividing by the modulus

$∥〈0,-12,-8〉∥=sqrt(0+144+64)=sqrt208=4sqrt13$

Thre unit vector is $=1/(4sqrt13)〈0,-12,-8〉$

$=〈0,-3/sqrt13,-2/sqrt13〉$

What is the unit vector that is orthogonal to the plane containing (3i + 2j - 3k) (3i + 2j - 3k) and (i -2j + 3k) (i -2j + 3k) ?