Question 65: Motion in 2D & Vectors

Question

Two vectors $$\overrightarrow X $$ and $$\overrightarrow Y $$ have equal magnitude. The magnitude of ($$\overrightarrow X $$ $$-$$ $$\overrightarrow Y $$) is n times the magnitude of ($$\overrightarrow X $$ + $$\overrightarrow Y $$). The angle between $$\overrightarrow X $$ and $$\overrightarrow Y $$ is :

Accepted Answer

Given X = Y
$$\sqrt {{X^2} + {Y^2} - 2 	imes Y\cos 	heta } $$
$$ = n\sqrt {{X^2} + {Y^2} + 2 	imes Y\cos 	heta } $$
Square both sides
$$2{X^2}(1 - \cos 	heta ) = {n^2}.2{X^2}(1 + \cos 	heta )$$
$$1 - \cos 	heta  = {n^2} + {n^2}\cos 	heta $$
$$\cos 	heta  = {{1 - {n^2}} \over {1 + {n^2}}}$$
$$	heta  = {\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over { - {n^2} - 1}}} ight]$$

Answer

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

Answer

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

Answer

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

Answer

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

Motion in 2D & Vectors

Two vectors $\overrightarrow X and \overrightarrow Y have equal magnitude. The magnitude of (\overrightarrow X - \overrightarrow Y ) is n times the magnitude of (\overrightarrow X + \overrightarrow Y ). The angle between \overrightarrow X and \overrightarrow Y $ is :

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