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The magnitude of two vectors P and Q differ by 1. The magnitude of their resultant makes an angle of \[{{\tan }^{-1}}\left(\dfrac{3}{4} \right)\] with P. The angle between P and Q is:
A. \[45{}^\circ \]
B. \[0{}^\circ \]
C. \[180{}^\circ \]
D. \[90{}^\circ \]

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Last updated date: 17th Jun 2024
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Answer
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Hint: In this question we have been asked to calculate the angle between vector P and vector Q. it is given that the magnitude of resultant vector makes an angle of \[{{\tan }^{-1}}\left( \dfrac{3}{4}\right)\] with vector P. Therefore, to solve this question, we shall use the triangle rule for vector to calculate the angle between vectors P and Q.
Formula used: -\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\]
Where,
C is the hypotenuse
A and B are opposite and adjacent sides.

Complete step-by-step solution:
 It is given that the magnitude of resultant vector P makes an angle of \[{{\tan }^{-1}}\left(\dfrac{3}{4} \right)\].
This means that the ratio of the two given vectors is \[\dfrac{3}{4}\]
Therefore,
 We know that,
\[\tan (x)=\dfrac{opposite}{adjacent}=\dfrac{P}{Q}\]
Now, by Pythagoras theorem,
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We know,
\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\]
After substituting properly,
\[R=\sqrt{{{3}^{2}}+{{4}^{2}}}\]
Therefore,
\[R=5\] …………………. (1)
Now, to calculate the angle between vector P and vector Q.
We know from triangle rule of vectors
\[R=\sqrt{{{P}^{2}}+{{Q}^{2}}+2PQ\cos \theta }\]
After substituting values and from (1)
We get,
\[{{R}^{2}}={{3}^{2}}+{{4}^{2}}+2\times 3\times 4\cos \theta \]
On solving,
We get
\[25=25+24\cos \theta \]
Therefore,
\[\theta ={{\cos }^{-1}}(0)\]
Therefore,
\[\theta =90{}^\circ \]
Therefore, the correct answer is option D.

Note: In mathematics, the Pythagorean theorem, also known as Pythagoras's theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides. In Pythagorean theorem, the hypotenuse is given by equation,
\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\].