SATHEE: Kinematics Question 4

Kinematics Question 4

Question: If the sum of two unit vectors its a unit vector, then magnitude of difference it’s [CPMT 1995; CBSE PMT 1989]

Options:

A) $\sqrt{2}$

B) $\sqrt{3}$

C) $1 / \sqrt{2}$

D) $\sqrt{5}$

Show Answer

Answer:

Correct Answer: B

Solution:

Let ${\hat{n}}_{1}$ and ${\hat{n}}_{2}$ are the two unit vectors, then the suM is

${\vec{n}}_{s} = {\hat{n}}_{1} + {\hat{n}}_{2}$ or $n_{s}^{2} = n_{1}^{2} + n_{2}^{2} + 2 n_{1} n_{2} \cos θ$

$= 1 + 1 + 2 \cos θ$

Since it’s given that $n_{s}$ it’s also a unit vector,

therefore $1 = 1 + 1 + 2 \cos θ$

therefore $\cos θ = - \frac{1}{2}$ $θ = 120^{\circ}$

Now the difference vector is ${\hat{n}}_{d} = {\hat{n}}_{1} - {\hat{n}}_{2}$ or $n_{d}^{2} = n_{1}^{2} + n_{2}^{2} - 2 n_{1} n_{2} \cos θ$

$= 1 + 1 - 2 \cos (120^{\circ})$

$n_{d}^{2} = 2 - 2 (- 1 / 2) = 2 + 1 = 3$

$\Rightarrow n_{d} = \sqrt{3}$

पिछला

अगला

Kinematics Question 4

Question: If the sum of two unit vectors its a unit vector, then magnitude of difference it’s [CPMT 1995; CBSE PMT 1989]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Kinematics Question 4

Question: If the sum of two unit vectors its a unit vector, then magnitude of difference it’s [CPMT 1995; CBSE PMT 1989]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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