Given two vectors a→ =2i^-3j^+k^ and b→=3i^+j^-4k^, find a unit vector n^ normal to the plane containing a→ and b→, such that a→, b→ and n→, in this order, obey the right –hand rule.
To find a unit vector n^ that is normal to the plane containing a→ and b→, we can take the cross product of a→ and b→:
n→ = a→ × b→
To find n→, we can use the following determinant:
|i j k|
|2 -3 1|
|3 1 -4|
By expanding this determinant, we get:
n→ = (3(-4) - 1(-3))i^ - (2(-4) - 1(1))j^ + (2(1) - 3(3))k^
= (-12 + 3)i^ - (-8 - 1)j^ + (2 - 9)k^
= -9i^ + 9j^ - 7k^
To find a unit vector n^, we divide n→ by its magnitude:
|n| = √((-9)^2 + (9)^2 + (-7)^2)
= √(81 + 81 + 49)
= √211
n→/|n| = (-9/√211)i^ + (9/√211)j^ + (-7/√211)k^
Therefore, a unit vector n^ that is normal to the plane containing a→ and b→, and obeys the right-hand rule with a→ and b→ is:
n^ = (-9/√211)i^ + (9/√211)j^ + (-7/√211)k^