How do you find a unit vector that is orthogonal to a and b where a = −7 i + 6 j − 8 k and b = −5 i + 6 j − 8 k?

Dec 28, 2016

The answer is =1/5〈0,-4,-3〉

Explanation:

To find a vector orthogonal to 2 other vectors, we must do a cross product.

The cross product of 2 vectors, veca=〈a,b,c〉 and vecb=〈d,e,f〉

is given by the determinant

$| \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(a , b , c\right) , \left(d , e , f\right) |$

$= \hat{i} | \left(b , c\right) , \left(e , f\right) | - \hat{j} | \left(a , c\right) , \left(d , f\right) | + \hat{k} | \left(a , b\right) , \left(d , e\right) |$

and $| \left(a , b\right) , \left(c , d\right) | = a d - b c$

Here, the 2 vectors are veca=〈-7,6,-8〉 and 〈-5,6,-8〉

And the cross product is

$| \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 7 , 6 , - 8\right) , \left(- 5 , 6 , - 8\right) |$

$= \hat{i} | \left(6 , - 8\right) , \left(6 , - 8\right) | - \hat{j} | \left(- 7 , - 8\right) , \left(- 5 , - 8\right) | + \hat{k} | \left(- 7 , 6\right) , \left(- 5 , 6\right) |$

$= \hat{i} \left(- 48 + 48\right) - \hat{i} \left(56 - 40\right) + \hat{k} \left(- 42 + 30\right)$

=〈0,-16,-12〉

Verification, by doing the dot product

〈0,-16,-12〉.〈-7,6,-8〉=0-96+96=0

〈0,-16,-12〉.〈-5,6,-8〉=0-96+96=0

Therefore, the vector is perpendicular to the other 2 vectors

The unit vector isobtained by dividing by the modulus.

The modulus is ∥〈0,-16,-12〉∥=sqrt(0+16^2+12^2)=sqrt400=20

The unit vector is =1/20〈0,-16,-12〉=1/5〈0,-4,-3〉