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

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Answer 1 · 2016-07-01T20:49:00

$= (-2 hat i + hat j + 7 hat k)/(3 sqrt(6))$

you'll do this by calculating the vector cross product of these 2 vectors to get the normal vector

so $vec n = (- 3 i + j -k) times (2i - 3 j +k )$

$= det [(hat i, hat j , hat k), (-3,1,-1),(2,-3,1)]$

$= hat i (1*1 -( -3*-1)) - hat j(-3*1 - (-1*2) ) + hat k (-3*-3 - 2*1))$

$= -2 hat i + hat j + 7 hat k$

the unit normal is $hat n = (-2 hat i + hat j + 7 hat k)/(sqrt((-2)^2 + 1^2 +7^2))$

$= (-2 hat i + hat j + 7 hat k)/(3 sqrt(6))$

you could check this by doing a scalar dot product between the normal and each of the original vectors, should get zero as they are orthogonal.

so for example

$vec v_1 * vec n$

$= (- 3 i + j -k) * (-2i + j + 7k )$

$= 6 + 1 - 7 = 0$

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