Question 72: Motion in 2D & Vectors

Question

Two vectors $${\overrightarrow P }$$ and $${\overrightarrow Q }$$ have equal magnitudes. If the magnitude of $${\overrightarrow P  + \overrightarrow Q }$$ is n times the magnitude of $${\overrightarrow P  - \overrightarrow Q }$$, then angle between $${\overrightarrow P }$$ and $${\overrightarrow Q }$$ is :

Accepted Answer

$$\left| {\overrightarrow P } ight| = \left| {\overrightarrow Q } ight| = x$$ ..... (i)
$$\left| {\overrightarrow P  + \overrightarrow Q } ight| = n\left| {\overrightarrow P  - \overrightarrow Q } ight|$$
$${P^2} + {Q^2} + 2PQ\cos 	heta  = {n^2}({P^2} + {Q^2} - 2PQ\cos 	heta )$$
Using (i) in above equation
$$\cos 	heta  = {{{n^2} - 1} \over {1 + {n^2}}}$$
$$	heta  = {\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} ight)$$

Answer

$${\sin ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)$$

Answer

$${\cos ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)$$

Answer

$${\sin ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$$

Answer

$${\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$$

Motion in 2D & Vectors

Two vectors ${\overrightarrow P } and {\overrightarrow Q } have equal magnitudes. If the magnitude of {\overrightarrow P + \overrightarrow Q } is n times the magnitude of {\overrightarrow P - \overrightarrow Q }, then angle between {\overrightarrow P } and {\overrightarrow Q }$ is :

Choose the correct answer: