Question

If a→=3i^-2j^-k^ and b→=i^+4j^+k^, find a unit vector n^ normal to the plane containing a→ and b→ such that a→, b→ and n^, in this order, form a right-handed system.​

Answer 1

The unit vector normal to the plane is ; n = $\frac{i}{\sqrt[3]{6} } - \frac{2j}{\sqrt[3]{6} } + \frac{7k}{\sqrt[3]{6} }$

Given data :

The unit vectors

a→=3i^-2j^-k^  and  b→=i^+4j^+k^

Determine the unit vector normal to the plane

The unit vector normal to the plane is calculated as

n = $\frac{a*b}{|b*b|}$  ---- ( 1 )

where ; a→ * b→ =  2i - 4j + 14k  and  |b * b| = $\sqrt{216}$  

back to equation ( 1 )

n = ( 2i - 4j + 14k ) / $\sqrt{216}$

  = $\frac{i}{\sqrt[3]{6} } - \frac{2j}{\sqrt[3]{6} } + \frac{7k}{\sqrt[3]{6} }$

Hence we can conclude that The unit vector normal to the plane is ; n = $\frac{i}{\sqrt[3]{6} } - \frac{2j}{\sqrt[3]{6} } + \frac{7k}{\sqrt[3]{6} }$

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