Question

A bird is flying with velocity (2i+3j-k) ms^-1 the angle made by the velocity with y axis is

Answer 1

Answer:

Cos^(-1) (3/√14) = theta

Explanation:

velocity of bird =(2i+3j-k)

longest diognal / hypotenous=√x^2+y^2+z^2 = √4+9+1= √14

therefore angle made by velocity with y axis

cos teta =3/√14

teta = cos^-1 (3/√14)

Answer 2

The angle made by the velocity with y axis is $cos (\frac{\sqrt{3} }{14} )$

Explanation:

Given:

A bird flying with velocity (2i + 3j + k ) m/s

To find: The angle made by velocity with y axis

Solution:

The velocity  is (2i + 3j + k ) m/s

We want to find the angle for the given vector makes with the y axis.

i.e  the unit vector j.

The product of the two unit vectors is the cosine of the angle between them.

The unit vector in the direction of A is $\frac{A}{|A|}$

⇒ $\frac{2i +3j +k}{|2i+3j+k|}$

= $\frac{2i +3j +k}{\sqrt{14} }$

= $\frac{2}{\sqrt{14} } i +\frac{3}{\sqrt{14} } j+\frac{1}{\sqrt{14} } k$

The cosine of the angle between the two vector is

⇒ $(\frac{2}{\sqrt{14} } i +\frac{3}{\sqrt{14} } j+\frac{1}{\sqrt{14} } k)j$  = $\frac{3}{\sqrt{14} }$

⇒ The angle that the vector $2i + 3j + k$ makes with the y axis is $cos (\frac{\sqrt{3} }{14} )$

Final answer:

The angle made by the velocity with y axis is $cos (\frac{\sqrt{3} }{14} )$

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