Question

a bird is flying with velocity(2i+3j-k)m/s, the angle made by velocity with y axis is​

Answer 1

Answer:

Cos^(-1) (3/√14) = theta

Explanation:

Concept is written...

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Answer 2

Answer:

The angle made by the velocity vector with the y-axis is​ $36.7$ °.

Explanation:

• When we calculate the dot product, two vectors are taken as input and the resulting output is scalar.
• The resulting scalar term depends on the angle between the vectors and the length of the individual vectors.

Step 1:

The velocity of the bird is $(2i+3j-k)$ $m/s$.

To calculate the angle made by the velocity with the y-axis, we apply the formula:

$A \cdot j= |A|\times |j|\times cos\alpha$

where $A$ is the vector

           $\alpha$ is the angle between the y-axis and the vector.

Step 2:

$(2i+3j-k) \cdot j= \sqrt{2^{2}+ 3^{2}+1^{2} } \times 1\times cos\alpha$

$3= \sqrt{14} \times 1\times cos\alpha$

$cos\alpha =\frac{3}{\sqrt{14} }$

$\alpha =cos^{-1} (\frac{3}{\sqrt{14} } )$

$\alpha =36.7$ °

So, the angle made by the velocity vector with the y-axis is​ $36.7$ °.