How would you find a unit vector perpendicular to plane ABC where points are A(3,-1,2), B(1,-1,-3) and C(4,-3,1)?

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Explanation

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Explanation:

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17
Sep 27, 2016

The desired unit vector is

$\left(- \frac{10}{\sqrt{165}} , - \frac{7}{\sqrt{165}} , - \frac{4}{\sqrt{165}}\right) .$

Explanation:

We know that, given two vectors, say vecx & vecy, their Vector

or Outer Product , denoted by $\vec{x} \times \vec{y} ,$ is a vector that is

perpendicular to the plane containing them.

The given pts. $A \left(3 , - 1 , 2\right) , B \left(1 , - 1 , - 3\right) \mathmr{and} C \left(4 , - 3 , 1\right)$ lie in the

plane $A B C$.

Accordingly, the vectors $\vec{A B} \mathmr{and} \vec{A C} \in \text{ the plane } A B C .$

Hence, $\vec{A B} \times \vec{A C} \bot \text{plane } A B C$.

Finally, the reqd. unit vector will be

$\frac{\vec{A B} \times \vec{A C}}{|} | \vec{A B} \times \vec{A C} | |$

We have, $\vec{A B} = \left(1 - 3 , - 1 + 1 , - 3 - 2\right) = \left(- 2 , 0 , - 5\right)$,

$\vec{A C} = \left(1 , - 2 , - 1\right)$, so that,

$\vec{A B} \times \vec{A C} = \det | \left(i , j , k\right) , \left(- 2 , 0 , - 5\right) , \left(1 , - 2 , - 1\right) |$

$= - 10 i - 7 j - 4 k = \left(- 10 , - 7 , - 4\right)$

$\Rightarrow | | \vec{A B} \times \vec{A C} | | = \sqrt{100 + 49 + 16} = \sqrt{165}$.

Finally, the desired unit vector is

$\left(- \frac{10}{\sqrt{165}} , - \frac{7}{\sqrt{165}} , - \frac{4}{\sqrt{165}}\right) .$

Enjoy Maths.!

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