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These results for vectors at right angle are exactly same as determined, using Pythagoras theorem.
Problem : Three radial vectors OA, OB and OC act at the center of a circle of radius “r” as shown in the figure. Find the magnitude of resultant vector.
Solution : It is evident that vectors are equal in magnitude and is equal to the radius of the circle. The magnitude of the resultant of horizontal and vertical vectors is :
$$\begin{array}{c}\mathrm{R\u2019}=\surd ({r}^{2}+{r}^{2})=\surd 2r\end{array}$$
The resultant of horizontal and vertical vectors is along the bisector of angle i.e. along the remaining third vector OB. Hence, magnitude of resultant of all three vectors is :
$$\begin{array}{c}\mathrm{R\u2019}=\mathrm{OB}+\mathrm{R\u2019}=r+\surd 2r=(1+\surd 2)r\end{array}$$
Problem : At what angle does two vectors a + b and a - b act so that the resultant is $\surd (3{a}^{2}+{b}^{2})$ .
Solution : The magnitude of resultant of two vectors is given by :
$$\begin{array}{c}R=\surd \{{(a+b)}^{2}+{(a-b)}^{2}+2(a+b\left)\right(a-b\left)\mathrm{cos}\theta \right\}\end{array}$$
Substituting the expression for magnitude of resultant as given,
$$\begin{array}{c}\Rightarrow \surd (3{a}^{2}+{b}^{2})=\surd \{{(a+b)}^{2}+{(a-b)}^{2}+2(a+b\left)\right(a-b\left)\mathrm{cos}\theta \right\}\end{array}$$
Squaring on both sides, we have :
$$\begin{array}{c}\Rightarrow (3{a}^{2}+{b}^{2})=\{{(a+b)}^{2}+{(a-b)}^{2}+2(a+b\left)\right(a-b\left)\mathrm{cos}\theta \right\}\end{array}$$
$$\begin{array}{c}\Rightarrow \mathrm{cos}\theta =\frac{({a}^{2}-{b}^{2})}{2({a}^{2}-{b}^{2})}=\frac{1}{2}=\mathrm{cos}60\xb0\end{array}$$
$$\begin{array}{c}\Rightarrow \theta =60\xb0\end{array}$$
The magnitude of sum of two vectors is either less than or equal to sum of the magnitudes of individual vectors. Symbolically, if a and b be two vectors, then
$$\begin{array}{c}|\mathbf{a}+\mathbf{b}|\le \left|\mathbf{a}\right|+\left|\mathbf{b}\right|\end{array}$$
We know that vectors a , b and their sum a + b is represented by three side of a triangle OAC. Further we know that a side of triangle is always less than the sum of remaining two sides. It means that :
$$\begin{array}{c}\mathrm{OC}<\mathrm{OA}+\mathrm{AC}\\ \mathrm{OC}<\mathrm{OA}+\mathrm{OB}\\ \Rightarrow |\mathbf{a}+\mathbf{b}|<\left|\mathbf{a}\right|+\left|\mathbf{b}\right|\end{array}$$
There is one possibility, however, that two vectors a and b are collinear and act in the same direction. In that case, magnitude of their resultant will be "equal to" the sum of the magnitudes of individual vector. This magnitude represents the maximum or greatest magnitude of two vectors being combined.
$$\begin{array}{c}\mathrm{OC}=\mathrm{OA}+\mathrm{OB}\\ \Rightarrow |\mathbf{a}+\mathbf{b}|=\left|\mathbf{a}\right|+\left|\mathbf{b}\right|\end{array}$$
Combining two results, we have :
$$\begin{array}{c}|\mathbf{a}+\mathbf{b}|\le \left|\mathbf{a}\right|+\left|\mathbf{b}\right|\end{array}$$
On the other hand, the magnitude of difference of two vectors is either greater than or equal to difference of the magnitudes of individual vectors. Symbolically, if a and b be two vectors, then
$$\begin{array}{c}|\mathbf{a}-\mathbf{b}|\ge \left|\mathbf{a}\right|-\left|\mathbf{b}\right|\end{array}$$
We know that vectors a , b and their difference a - b are represented by three side of a triangle OAE. Further we know that a side of triangle is always less than the sum of remaining two sides. It means that sum of two sides is greater than the third side :
$$\begin{array}{c}\mathrm{OE}+\mathrm{AE}>\mathrm{OA}\\ \Rightarrow \mathrm{OE}>\mathrm{OA}-\mathrm{AE}\\ \Rightarrow |\mathbf{a}-\mathbf{b}|>\left|\mathbf{a}\right|-\left|\mathbf{b}\right|\end{array}$$
There is one possibility, however, that two vectors a and b are collinear and act in the opposite directions. In that case, magnitude of their difference will be equal to the difference of the magnitudes of individual vector. This magnitude represents the minimum or least magnitude of two vectors being combined.
$$\begin{array}{c}\Rightarrow \mathrm{OE}=\mathrm{OA}-\mathrm{AE}\\ \Rightarrow |\mathbf{a}-\mathbf{b}|=\left|\mathbf{a}\right|-\left|\mathbf{b}\right|\end{array}$$
Combining two results, we have :
$$\begin{array}{c}|\mathbf{a}-\mathbf{b}|\ge \left|\mathbf{a}\right|-\left|\mathbf{b}\right|\end{array}$$
Lami's theorem relates magnitude of three non-collinear vectors with the angles enclosed between pair of two vectors, provided resultant of three vectors is zero (null vector). This theorem is a manifestation of triangle law of addition. According to this theorem, if resultant of three vectors a , b and c is zero (null vector), then
$$\begin{array}{c}\frac{a}{\mathrm{sin}\alpha}=\frac{b}{\mathrm{sin}\beta}=\frac{c}{\mathrm{sin}\gamma}\end{array}$$
where α, β and γ be the angle between the remaining pairs of vectors.
We know that if the resultant of three vectors is zero, then they are represented by three sides of a triangle in magnitude and direction.
Considering the magnitude of vectors and applying sine law of triangle, we have :
$$\begin{array}{c}\frac{\mathrm{AB}}{\mathrm{sin}\mathrm{BCA}}=\frac{\mathrm{BC}}{\mathrm{sin}\mathrm{CAB}}=\frac{\mathrm{CA}}{\mathrm{sin}\mathrm{ABC}}\end{array}$$
$$\begin{array}{c}\Rightarrow \frac{\mathrm{AB}}{\mathrm{sin}(\pi -\alpha )}=\frac{\mathrm{BC}}{\mathrm{sin}(\pi -\beta )}=\frac{\mathrm{CA}}{\mathrm{sin}(\pi -\gamma )}\end{array}$$
$$\begin{array}{c}\Rightarrow \frac{\mathrm{AB}}{\mathrm{sin}\alpha}=\frac{\mathrm{BC}}{\mathrm{sin}\beta}=\frac{\mathrm{CA}}{\mathrm{sin}\gamma}\end{array}$$
It is important to note that the ratio involves exterior (outside) angles – not the interior angles of the triangle. Also, the angle associated with the magnitude of a vector in the individual ratio is the included angle between the remaining vectors.
Two forces of 10 N and 25 N are applied on a body. Find the magnitude of maximum and minimum resultant force.
Resultant force is maximum when force vectors act along the same direction. The magnitude of resultant force under this condition is :
$$\begin{array}{c}{R}_{\text{max}}=10+25=35\phantom{\rule{2pt}{0ex}}N\\ {R}_{\text{min}}=25-10=15\phantom{\rule{2pt}{0ex}}N\end{array}$$
Can a body subjected to three coplanar forces 5 N, 17 N and 9 N be in equilibrium?
The resultant force, on a body in equilibrium, is zero. It means that three forces can be represented along three sides of a triangle. However, we know that sum of any two sides is greater than third side. In this case, we see that :
$$\begin{array}{c}5+9<17\end{array}$$
Clearly, three given forces can not be represented by three sides of a triangle. Thus, we conclude that the body is not in equilibrium.
Under what condition does the magnitude of the resultant of two vectors of equal magnitude, is equal in magnitude to either of two equal vectors?
We know that resultant of two vectors is represented by the closing side of a triangle. If the triangle is equilateral then all three sides are equal. As such magnitude of the resultant of two vectors is equal to the magnitude of either of the two vectors.
Under this condition, vectors of equal magnitude make an angle of 120° between them.
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