Vectors

The angle is $120^\circ$.

For two equal forces $F$ at angle $\theta$, $R^2=F^2+F^2+2F^2\cos\theta=2F^2(1+\cos\theta)$. Setting $R=F$:

$F^2=2F^2(1+\cos\theta)\;\Rightarrow\;\cos\theta=-\tfrac12\;\Rightarrow\;\theta=120^\circ$

Check: the two forces and their resultant then form an equilateral triangle, with all three sides equal to $F$.

Relative velocity of B with respect to A is

$\vec V_{BA}=\vec V_B-\vec V_A=(3\hat i+2\hat j-\hat k)-(\hat i+2\hat j-3\hat k)=2\hat i+0\hat j+2\hat k$

So $\vec V_{BA}=2\hat i+2\hat k$, with magnitude $\sqrt{2^2+2^2}=2\sqrt2\approx2.83$ units.

Take north as $+y$ and east as $+x$.

Net northward displacement: $3.25-1.50=1.75\ \text{km}$. Net westward displacement: $4.75\ \text{km}$.

Magnitude:

$R=\sqrt{(1.75)^2+(4.75)^2}=\sqrt{3.0625+22.5625}=\sqrt{25.625}\approx 5.06\ \text{km}$

Direction:

$\tan\theta=\frac{1.75}{4.75}=0.368\;\Rightarrow\;\theta\approx 20.2^\circ$

The resultant displacement is about $5.06\ \text{km}$, directed $20.2^\circ$ north of west.

(c) $90^\circ$

Squaring both sides of $|\vec A+\vec B|=|\vec A-\vec B|$ gives $A^2+B^2+2AB\cos\theta=A^2+B^2-2AB\cos\theta$, so $4AB\cos\theta=0$. Since $A,B\neq0$, $\cos\theta=0$ and $\theta=90^\circ$.

Let a vector $\vec A$ make an angle $\theta$ with the $x$-axis. Draw perpendiculars from its head to the $x$- and $y$-axes. This gives two mutually perpendicular vectors $\vec A_x$ and $\vec A_y$ whose vector sum is $\vec A$, so they are the rectangular components of $\vec A$.

From the right-angled triangle formed:

$A_x = A\cos\theta,\qquad A_y = A\sin\theta$

Together they give back the vector, since

$A=\sqrt{A_x^2+A_y^2},\qquad \tan\theta=\frac{A_y}{A_x}$

In unit-vector form, $\vec A = A\cos\theta\,\hat i + A\sin\theta\,\hat j$.

(b) energy

Energy is a scalar, having magnitude only. Torque, angular momentum and electric field intensity all have direction as well as magnitude.

The resultant is zero when the two vectors have equal magnitudes and act in exactly opposite directions (angle $180^\circ$ between them), so that $\vec A+\vec B=0$.

(b) 5 units

Since the vectors are at right angles,
$R=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\ \text{units}$

Parallelogram law of vector addition

If two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a common point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that same point.

Derivation of the magnitude of the resultant

Let the two vectors $\vec{A}$ and $\vec{B}$ be inclined at an angle $\theta$. Draw them as the adjacent sides $OA$ and $OB$ of a parallelogram $OACB$, with $OA = A$ and $OB = B$. By the law, the resultant $\vec{R}$ is the diagonal $OC$.

Since $AC$ is parallel and equal to $OB$, we have $AC = B$, and the angle between $OA$ and $AC$ is $180^\circ - \theta$ (because $AC$ is parallel to $OB$ but points in the opposite sense along the line).

Now apply the law of cosines to triangle $OAC$:

$OC^2 = OA^2 + AC^2 - 2(OA)(AC)\cos(180^\circ - \theta)$

Since $\cos(180^\circ - \theta) = -\cos\theta$,

$OC^2 = A^2 + B^2 - 2AB(-\cos\theta)$

$OC^2 = A^2 + B^2 + 2AB\cos\theta$

Therefore, the magnitude of the resultant is

$R = |\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}$

Special cases

• If $\theta = 0^\circ$ (vectors parallel), $R = \sqrt{A^2 + B^2 + 2AB} = A + B$ (maximum).
• If $\theta = 90^\circ$ (vectors perpendicular), $R = \sqrt{A^2 + B^2}$.
• If $\theta = 180^\circ$ (vectors antiparallel), $R = \sqrt{A^2 + B^2 - 2AB} = A - B$ (minimum).

The direction of the resultant is given by $\tan\alpha = \dfrac{B\sin\theta}{A + B\cos\theta}$, where $\alpha$ is the angle the resultant makes with $\vec{A}$.

Scalar product (dot product): The scalar product of two vectors $\vec{A}$ and $\vec{B}$ is defined as the product of their magnitudes and the cosine of the angle $\theta$ between them: $\vec{A} \cdot \vec{B} = AB\cos\theta$ The result is a scalar quantity.

Vector product (cross product): The vector product of two vectors $\vec{A}$ and $\vec{B}$ is defined as a vector whose magnitude is the product of their magnitudes and the sine of the angle $\theta$ between them, and whose direction is perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ (given by the right-hand rule): $\vec{A} \times \vec{B} = AB\sin\theta\,\hat{n}$ The result is a vector quantity.

Differences:

Volume of the parallelepiped:

Let the three coterminous edges of a parallelepiped be represented by $\vec{A}$, $\vec{B}$, and $\vec{C}$.

The vector product $\vec{B}\times\vec{C}$ has magnitude $|\vec{B}\times\vec{C}| = BC\sin\theta$ which is numerically equal to the area of the parallelogram formed by $\vec{B}$ and $\vec{C}$ (the base of the parallelepiped). Its direction is perpendicular to the plane of $\vec{B}$ and $\vec{C}$, i.e. along the normal to the base.

Now $\vec{A}\cdot(\vec{B}\times\vec{C}) = |\vec{A}|\,|\vec{B}\times\vec{C}|\cos\phi$, where $\phi$ is the angle between $\vec{A}$ and the normal to the base.

But $|\vec{A}|\cos\phi$ is the perpendicular height $h$ of the parallelepiped, and $|\vec{B}\times\vec{C}|$ is the area of the base $S$. Therefore, $\vec{A}\cdot(\vec{B}\times\vec{C}) = S \times h = \text{volume of the parallelepiped}$

Hence the scalar triple product $\vec{A}\cdot(\vec{B}\times\vec{C})$ is equal in magnitude to the volume of the parallelepiped formed by $\vec{A}$, $\vec{B}$, and $\vec{C}$. (It is also written as $[\vec{A}\ \vec{B}\ \vec{C}]$.)

Let the projectile be projected from the origin (O) with initial velocity (u) at an angle (\theta) with the horizontal. Take the horizontal direction as the (x)-axis and the vertical upward direction as the (y)-axis.

Resolve the initial velocity into components:

[ u_x = u\cos\theta, \qquad u_y = u\sin\theta ]

Along the horizontal direction, acceleration (a_x=0).
Along the vertical direction, acceleration (a_y=-g).

(i) Time of flight

Let (T) be the total time of flight. The vertical displacement at the instant of landing is zero:

[ y = u_y T - \frac12 gT^2 = 0 ]

[ u\sin\theta,T - \frac12 gT^2 = 0 ]

[ T\left(u\sin\theta - \frac12 gT\right)=0 ]

Since (T\neq 0),

[ u\sin\theta = \frac12 gT ]

[ \boxed{T = \frac{2u\sin\theta}{g}} ]

(ii) Maximum height

At the highest point, the vertical component of velocity becomes zero. Let the time taken to reach the highest point be (t).

[ v_y = u\sin\theta - gt = 0 ]

[ t = \frac{u\sin\theta}{g} ]

The maximum height (H) is the vertical displacement in this time:

[ H = u\sin\theta \cdot t - \frac12 gt^2 ]

[ H = u\sin\theta\left(\frac{u\sin\theta}{g}\right)

• \frac12 g\left(\frac{u\sin\theta}{g}\right)^2 ]

[ H = \frac{u^2\sin^2\theta}{g} - \frac{u^2\sin^2\theta}{2g} ]

[ \boxed{H = \frac{u^2\sin^2\theta}{2g}} ]

(iii) Horizontal range

The horizontal range is the horizontal distance covered during the total time of flight (T).

[ R = u_x T ]

[ R = u\cos\theta \cdot \frac{2u\sin\theta}{g} ]

[ R = \frac{u^2(2\sin\theta\cos\theta)}{g} ]

Using (2\sin\theta\cos\theta = \sin 2\theta),

[ \boxed{R = \frac{u^2\sin 2\theta}{g}} ]

Condition for maximum horizontal range

From the expression,

[ R = \frac{u^2\sin 2\theta}{g} ]

For a given initial speed (u), the range is maximum when (\sin 2\theta) is maximum. The maximum value of sine is (1), so

[ \sin 2\theta = 1 ]

[ 2\theta = 90^\circ ]

[ \boxed{\theta = 45^\circ} ]

Thus, the horizontal range is maximum when the projectile is fired at (45^\circ) with the horizontal.

The maximum horizontal range is obtained by putting (\theta=45^\circ):

[ R_{\max} = \frac{u^2\sin(2\times45^\circ)}{g} ]

[ R_{\max} = \frac{u^2\sin90^\circ}{g} ]

Since (\sin90^\circ=1),

[ \boxed{R_{\max} = \frac{u^2}{g}} ]

So the maximum horizontal range is (\boxed{\dfrac{u^2}{g}}).