The magnitude of resultant of the two vectors of magnitude $4$ and $3$ is $1$ . The angle between them is:

z e r o

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\frac{π}{4}

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\frac{π}{2}

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π

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Solution

The correct option is A $π$
Given two vector A=4 and B=3 and the resultant is 1.
We have the formula $R = \sqrt{A^{2} + B^{2} - 2 \times A . B . c o s θ}$
So, $1 = \sqrt{4^{2} + 3^{2} - 2 \times 4 \times 3 \times c o s θ}$
$\pm 1 = 25 - 24 c o s θ$
Taking +1 we get:-
$c o s θ = \frac{24}{24}$
$c o s θ = 1$
$θ = π$

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⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

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Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is 5 unit

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