# Find a unit vector in the plane of i+2j,j+2k perpendicular to 2i+j+2k?

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Feb 9, 2018

$\setminus$

$\text{Answer:} \setminus q \quad \setminus q \quad \frac{1}{5 \setminus \sqrt{5}} \left(5 i + 6 j - 8 k\right) .$

#### Explanation:

$\setminus$

 "Let:" \qquad A = i+2j, \quad B = j+2k; \quad \ \ L =2i+j+2k.

$\text{Let:" \qquad P = "the plane formed by" \ \ A \ \ "and} \setminus \setminus B .$

$\text{Here's the idea:}$

1. $N \setminus = \setminus A \times B \setminus \setminus \text{will be a vector normal to the plane} \setminus \setminus P .$
2. $D \setminus = \setminus L \times N \setminus \setminus \text{will be a vector perpendicular to both" \ L \ "&} \setminus N$
3. $\text{Because" \ D \ \ "is perpendicular to" \ N, "and" \ N \ "is normal to} \setminus P ,$
$\text{we have" \ D \ \ "is in the plane} \setminus P .$
4. $\text{So" \ D \ "will be both in the plane" \ P, "and perpendicular to} \setminus L , \setminus$
$\text{by (2)} .$
5. $\text{So, after normalizing" \ D \ "to a unit vector} \setminus \setminus \hat{D} ,$
$\setminus \hat{D} \setminus \text{will have all the desired properties, and}$
$\text{will be our answer.}$

$\text{Now we compute" \ D, \ "and then normalize it.}$

$\text{Following the definitons of the vectors above, we have:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad D \setminus = \setminus L \times N \setminus = \setminus L \times \left(A \times B\right) .$

$\text{Thus:}$

$D \setminus = \setminus L \times \left(A \times B\right) \setminus = \setminus L \times \left(\left(i + 2 j\right) \times \left(j + 2 k\right)\right)$

$\setminus \quad \setminus \quad = \setminus L \times | \left(i , j , k\right) , \left(1 , 2 , 0\right) , \left(0 , 1 , 2\right) |$

$\setminus \quad \setminus \quad = \setminus L \times \left(| \left(2 , 0\right) , \left(1 , 2\right) | i - | \left(1 , 0\right) , \left(0 , 2\right) | j + | \left(1 , 2\right) , \left(0 , 1\right) | k\right)$

$\setminus \quad \setminus \quad = \setminus L \times \left(4 i - 2 j + k\right)$

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

$\setminus \quad \setminus \quad = \setminus | \left(i , j , k\right) , \left(2 , 1 , 2\right) , \left(4 , - 2 , 1\right) |$

$\setminus \quad \setminus \quad = \setminus \left(| \left(1 , 2\right) , \left(- 2 , 1\right) | i - | \left(2 , 2\right) , \left(4 , 1\right) | j + | \left(2 , 1\right) , \left(4 , - 2\right) | k\right)$

$\setminus \quad \setminus \quad = \setminus \left(5 i + 6 j - 8 k\right) .$

$\text{Hence:} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad D = 5 i + 6 j - 8 k .$

$\text{Recalling (4) above:}$

$\setminus q \quad \setminus \quad D \setminus \text{will be both in the plane" \ P, "and perpendicular to} \setminus L .$

$\setminus q \quad \setminus q \quad \therefore \setminus \quad \setminus D \setminus \text{is what we want, after normalizing it to} \setminus \setminus \hat{D} .$

$\text{Normalization of" \ D \ "to} \setminus \setminus \hat{D} :$

$\text{Hence:} \setminus q \quad \setminus \hat{D} = \frac{D}{|} | D | | \setminus = \setminus \frac{5 i + 6 j - 8 k}{\setminus} \sqrt{{5}^{2} + {6}^{2} + {\left(- 8\right)}^{2}}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad = \setminus \frac{5 i + 6 j - 8 k}{\setminus} \sqrt{125} \setminus = \setminus \frac{1}{5 \setminus \sqrt{5}} \left(5 i + 6 j - 8 k\right) .$

$\text{So we have the desired vector": } \setminus \quad \setminus \quad \setminus \hat{D} \setminus = \setminus \frac{1}{5 \setminus \sqrt{5}} \left(5 i + 6 j - 8 k\right) .$

$\setminus$

$\text{Summarizing:}$

$\text{Solution vector": } \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \quad \setminus \hat{D} \setminus = \setminus \frac{1}{5 \setminus \sqrt{5}} \left(5 i + 6 j - 8 k\right) .$

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