# What is the unit vector that is normal to the plane containing (3i + 4j - k) and (2i+ j - 3k)?

$= \setminus \pm \left(\setminus \frac{- 11 i + 7 j - 5 k}{\setminus \sqrt{195}}\right)$

#### Explanation:

Taking the cross product of given vectors $\left(3 i + 4 j - k\right)$ & $\left(2 i + j - 3 k\right)$ as follows

$\left(3 i + 4 j - k\right) \setminus \times \left(2 i + j - 3 k\right)$

$= - 11 i + 7 j - 5 k$

Above vector will be normal to the given vectors hence the unit normal vector

$\setminus \hat{n} = \setminus \frac{- 11 i + 7 j - 5 k}{| - 11 i + 7 j - 5 k |}$

$= \setminus \frac{- 11 i + 7 j - 5 k}{\setminus \sqrt{{\left(- 11\right)}^{2} + {7}^{2} + {\left(- 5\right)}^{2}}}$

$= \setminus \frac{- 11 i + 7 j - 5 k}{\setminus \sqrt{195}}$

hence the unit vector $\setminus \hat{n}$ normal to the plane containing given vectors will be

$= \setminus \pm \left(\setminus \frac{- 11 i + 7 j - 5 k}{\setminus \sqrt{195}}\right)$

Jul 28, 2018

The unit vector is = <〈-11/sqrt195,7/sqrt195,-5/sqrt195〉>

#### Explanation:

The vector normal to a plane containing $2$ vectors is given by the cross product of 2 vectors, This is calculated with the determinant

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(d , e , f\right) , \left(g , h , i\right) |$

where veca=〈d,e,f〉 and vecb=〈g,h,i〉 are the 2 vectors

Here, we have veca=〈3,4,-1〉 and vecb=〈2,1,-3〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(3 , 4 , - 1\right) , \left(2 , 1 , - 3\right) |$

$= \vec{i} | \left(4 , - 1\right) , \left(1 , - 3\right) | - \vec{j} | \left(3 , - 1\right) , \left(2 , - 3\right) | + \vec{k} | \left(3 , 4\right) , \left(2 , 1\right) |$

$= \vec{i} \left(\left(4\right) \cdot \left(- 3\right) - \left(1\right) \cdot \left(- 1\right)\right) - \vec{j} \left(\left(3\right) \cdot \left(- 3\right) - \left(2\right) \cdot \left(- 1\right)\right) + \vec{k} \left(\left(3\right) \cdot \left(1\right) - \left(4\right) \cdot \left(2\right)\right)$

=〈-11,7,-5〉=vecc

Verification by doing 2 dot products

〈-11,7,-5〉.〈3,4,-1〉=(-11)*(3)+(7)*(4)+(-5)*(-1)=0

〈-11,7,-5〉.〈2,1,-3〉=(-11)*(2)+(7)*(1)+(-5)*(-3)=0

So,

$\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$

||vecc||= ||〈-11,7,-5〉|| = sqrt((-11)^2+(7)^2+(-5)^2)

$= \sqrt{121 + 49 + 25}$

$= \sqrt{195}$

The unit vector is

hatc=vecc/(||vecc||)=1/sqrt(195)〈-11,7,-5〉