Question

What will be the unit vector in the direction of (3i+j+2k)?

Answer 1

Answer:(3i+1j+2k)/14

Step-by-step explanation:

Magnitude=√3^2+1^2+2^2=14

Unit vector =(3i+1j+2k)/14

Answer 2

Answer:

The unit vector in the direction of (3i+j+2k) = $\frac{3i+j+2k}{\sqrt{14} }$

Step-by-step explanation:

Required to find,

The unit vector in the direction of (3i+j+2k)

Recall the formula

Unit vector = $\frac{Vector}{Magnitude \ of \ the \ vector}$

The magnitude of the vector ai+bj+ck is =  $\sqrt{(a^2+b^2+c^2)}$

Solution:

The given vector is 3i+j+2k

The magnitude of the vector = $\sqrt{(3^2+1^2+2^2)}$ = $\sqrt{(9+1+4)}$ = $\sqrt{14}$

Unit vector = $\frac{Vector}{Magnitude \ of \ the \ vector}$ = $\frac{3i+j+2k}{\sqrt{14} }$

∴ Unit vector  in the direction of (3i+j+2k) = $\frac{3i+j+2k}{\sqrt{14} }$

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