Question #178443

Vector A has magnitude 4 units, vector B has magnitude 6 units. The angle between A and B is 60 degrees. Calculate the magnitude of 2A + 3B.


Expert's answer

Explanations & Calculations


  • What is asked is the vector addition of 2 vectors and the included angle is given.
  • In such a case, using the equation R2=P2+Q2+2PQcos⁡θ\small R^2=P^2+Q^2+2PQ\cos\theta, the magnitude of their resultant (R) can be calculated.


  • Now the two forces are P⃗=2A⃗\small \vec{P}=2\vec{A} and Q⃗=3B⃗\small \vec{Q}=3\vec{B}
  • Then the corresponding magnitudes are

∣P⃗∣=P=∣2A⃗∣=2∣A⃗∣=2×4units=8 units\qquad\small |\vec{P}|=P=|2\vec{A}|=2|\vec{A}|=2\times4units=8\,units

∣Q⃗∣=Q=∣3Q⃗∣=3∣Q⃗∣=3×6 units=18 units.\qquad\small |\vec{Q}|=Q=|3\vec{Q}|=3|\vec{Q}|=3\times6\,units=18\,units.


  • Then plugging these data in the above equation,

R2=82+182+2(8)(18)cos⁡60=532 units2R=23.065N\qquad\qquad \begin{aligned} \small R^2&=\small8^2+18^2+2(8)(18)\cos60\\ &=\small 532\,units^2 \\ \small R &=\small \bold{23.065N} \end{aligned}


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