SATHEE: Motion in a Plane - Result Question 6

Motion in a Plane - Result Question 6

7. $\vec{A}$ and $\vec{B}$ are two vectors and $θ$ is the angle between them, if $| \vec{A} \times \vec{B} | = \sqrt{3} (\vec{A} \cdot \vec{B})$ , the value of $θ$ is

(a) $45^{\circ}$

(b) $30^{\circ}$

(d) $60^{\circ}$

[2007]

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Answer:

Correct Answer: 7. (d)

Solution:

(d) $| \vec{A} \times \vec{B} | = \sqrt{3} (\vec{A} \cdot \vec{B})$

$\Rightarrow A B \sin θ = \sqrt{3} A B \cos θ$

$\Rightarrow \tan θ = \sqrt{3} \Rightarrow θ = 60^{\circ}$

(d) $| \vec{A} + \vec{B} |^{2} = | \vec{A} - \vec{B} |^{2}$

$= | \vec{A} |^{2} + | \vec{B} |^{2} + 2 \vec{A} \cdot \vec{B} = A^{2} + B^{2} + 2 A B \cos θ$

$= | \vec{A} - \vec{B} |^{2} = | \vec{A} |^{2} + | \vec{B} |^{2} - 2 \vec{A} \cdot \vec{B}$

$= A^{2} + B^{2} - 2 A B \cos θ$

So, $A^{2} + B^{2} + 2 A B \cos θ$

$= A^{2} + B^{2} - 2 A B \cos θ$

$4 A B \cos θ = 0 \Rightarrow \cos θ = 0$

$θ = 90^{\circ}$

So, angle between $A$ & $B$ is $90^{\circ}$ .

Motion in a Plane - Result Question 6

7. $\vec{A}$ and $\vec{B}$ are two vectors and $θ$ is the angle between them, if $| \vec{A} \times \vec{B} | = \sqrt{3} (\vec{A} \cdot \vec{B})$ , the value of $θ$ is

Answer:

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Motion in a Plane - Result Question 6

7. A→ and B→ are two vectors and θ is the angle between them, if |A→×B→|=3(A→⋅B→), the value of θ is

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7. $\vec{A}$ and $\vec{B}$ are two vectors and $θ$ is the angle between them, if $| \vec{A} \times \vec{B} | = \sqrt{3} (\vec{A} \cdot \vec{B})$ , the value of $θ$ is