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91Ó°ÊÓ

Find a unit vector normal to the plane containing

v→=2i^+3j^-k^w→=-2i^-4j^+3k^

Short Answer

Expert verified

a^=-13237i^+8237j^-2237k^

Step by step solution

01

Step 1. Given vectors in the plane 

v→=2i^+3j^-k^w→=-2i^-4j^+3k^

02

Step 2. The cross product

We know that the unit vector perpendicular to

is:

a^=±v→×w→v→×w→

So;

v→=2i^+3j^-k^w→=-2i^-4j^-3k^v→×w→=i^j^k^23-1-2-4-3v→×w→=i^(-9-4)-j^(-6-2)+k^(-8+6)v→×w→=-13i^+8j^-2k^|v→×w→|=169+4+64|v→×w→|=237

03

Step 3. The unit vector is :

a^=-13237i^+8237j^-2237k^

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