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Two proper vectors $ \bar a $ and $ \bar b $ are connected by $ \bar a + \bar b = - \bar b $ . The ratio of the magnitudes of the vectors $ \bar a $ and $ \bar b $ is
(A) $ 1:1 $
(B) $ 2:1 $
(C) $ 1:2 $
(D) $ 3:1 $

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint The magnitude of a vector ignores the direction, thus, the sign of the vector. The ratio of two numbers A and B is given as $ {\text{A:B}} $ . So we need to calculate the magnitude of one vector in terms of the other to get the ratio.

Complete step by step answer
From the question, we have two proper vectors $ \bar a $ and $ \bar b $ to be connected by the expression
 $ \Rightarrow \bar a + \bar b = - \bar b $
To find the ratio of the magnitudes, we must first find the value or expression for the vector $ \bar a $ .
Hence, calculating the vector $ \bar a $ from the connection between the vectors above, we subtract the vector $ \bar b $ from both sides of the equation. Thus we have that
 $ \Rightarrow \bar a = - \bar b - \bar b = - 2\bar b $
 $ \therefore \bar a = - 2\bar b $ . Obviously, this implies that the vector $ \bar a $ is twice the negative of the vector $ \bar b $ .
However, we are only to calculate the ratio of the magnitudes, hence we must first find the magnitude of the vector $ \bar a $ .
The magnitude of the vector is given as
 $ \Rightarrow \left| {\bar a} \right| = 2\bar b $ . In calculating magnitudes, the direction of the vector is not considered. This reflects by ignoring the sign of the vector.
Now the ratio of the magnitudes of $ \bar a $ and $ \bar b $ can be given as
 $ \Rightarrow \left| {\bar a} \right|:\left| {\bar b} \right| $ . Since the magnitude of $ \bar a $ is $ \left| {\bar a} \right| = 2\bar b $ , then
 $ \Rightarrow \left| {\bar a} \right|:\left| {\bar b} \right| = 2:1 $
We can cancel the vector $ \bar b $ , hence,
 $ \Rightarrow \left| {\bar a} \right|:\left| {\bar b} \right| = 2:1 $
Hence, the correct answer is option B.

Note
Alternatively, to obtain the magnitudes, we can also express vector $ \bar b $ in terms of $ \bar a $ as in:
 $ \bar a = - 2\bar b $
 $ \Rightarrow \bar b = - \dfrac{{\bar a}}{2} $ . Following the same path. By finding the magnitude of the vector $ \bar b $ . Hence, we have that
 $ \left| {\bar b} \right| = \dfrac{{\bar a}}{2} $ . Thus finding the ratio of the magnitudes, we have that
 $ \left| {\bar a} \right|:\left| {\bar b} \right| = \bar a:\dfrac{{\bar a}}{2} $ . Similarly, by eliminating the vector, we have
 $ \left| {\bar a} \right|:\left| {\bar b} \right| = 1:\dfrac{1}{2} $ . Multiplying by 2 we have
 $ \left| {\bar a} \right|:\left| {\bar b} \right| = 2:1 $ which is identical to the answer above.
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