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91Ó°ÊÓ

Find a unit vector normal to the plane containing

v→=i^+3j^-2k^w→=-2i^+j^+3k^

Short Answer

Expert verified

a^=11i^+j^+7k^171

Step by step solution

01

Step 1. Given vectors in the plane 

v→=i^+3j^-2k^w→=-2i^+j^+3k^

02

Step 2. The cross product

We know that the unit vector perpendicular to v→&w→

is :role="math" localid="1646870930687" a^=±v→×w→v→×w→

So;

v→=i^+3j^-2k^w→=-2i^+j^+3k^v→×w→=i^j^k^13-2-213v→×w→=i^(9+2)-j^(3-4)+k^(1+6)v→×w→=11i^+j^+7k^|v→×w→|=121+1+49|v→×w→|=171

03

Step 3. The unit vector is :

a^=11i^+j^+7k^171

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