Question

the magnitude of vectors p and q differ by 1 . the magnitude of their resultant makes an angle of tan^-1(3/4) with p . the an gle between p and q is

Answer 1

$\huge{\boxed{\boxed{\red{A}{N}{\blue{S}{\red{W}{\green{E}{R}}}}}}}$

Tan = $\tt{\boxed{\frac{3}{4}}}$

Perpendicular = $\tt{3}$

Base = $\tt{3 + 1 = 4}$

Hypotenuse = $\tt{\sqrt{ {3}^{2} + {4}^{2} }}$

Now, to find the Angle between P and Q we have to use the Formula of Triangle law of Vector.

• $\tt{{5}^{2} = \sqrt{ {p}^{2} + {q}^{2} \: + 2pq \: cos(x) }}$
• $\tt{x = \angle \: between \: p \: and \: q}$
• $\tt{ 25 = 25 + 2pq\: cos(x)}$
• $\tt{ 25 = 25 + 2 \times 3 \times 4\: cos(x)}$
• $\tt{ 25 = 25 + 24\: cos(x)}$
• $\tt{cos(x) = 0}$

$\tt{\implies{\huge{\boxed{\boxed{x = 90°}}}}}$

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