**Establish the following vector inequalities geometrically or otherwise:**

**(a) \(\left | a + b \right |\leq \left | a \right | + \left | b \right |\)**

**(b) \(|a-b|\geq||a|-|b||\)**

**(c) \(\left | a – b \right |\leq \left | a \right | + \left | b \right |\)**

**(d) \(| a – b |\geq | | a | – | b | |\)**

**When does the equality sign above apply?**

Asked by Pragya Singh | 1 year ago | 109

**(a)** Let two vectors \(\vec{a}\;and \;\vec{b}\)

be represented by the adjacent sides of a parallelogram PQRS,

Here,

\( \left | \vec{QR} \right | = \left | \vec{a} \right |\)........(i)

\( \left | \vec{RS} \right | = \left | \vec{QP} \right | = \left | \vec{b} \right |\)..........(ii)

\( \left | \vec{QS} \right | = \left | \vec{a} + \vec{b} \right |\) ............(iii)

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in \(\Delta QRS,\)

QS < (QR + RS)

\( \left | \vec{a} + \vec{b} \right | < \left | \vec{a} \right | + \left | \vec{b} \right |\) ........(iv)

If the two vectors \(\vec{a}\) and \( \vec{b}\)act along a straight line in the same direction, then:

\(\left | \vec{a} + \vec{b} \right | = \left | \vec{a} \right | + \left | \vec{b} \right |\) ...........(v)

**Combine equation (iv) and (v),**

**\(\left | \vec{a} + \vec{b} \right | \leq \left | \vec{a} \right | + \left | \vec{b} \right |\)**

**(b)** Let two vectors \(\vec{a}\;and \;\vec{b}\) be represented by the adjacent sides of a** **parallelogram PQRS**,** as given in the figure.

Here,

\(\left | \vec{QR} \right | = \left | \vec{a} \right |\) ........(i)

\(\left | \vec{RS} \right | = \left | \vec{QP} \right | = \left | \vec{b} \right |\) ..........(ii)

\(\left | \vec{QS} \right | = \left | \vec{a} + \vec{b} \right |\) ..........(iii)

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in \(\Delta QRS\),

QS + RS > QR

QS + QR > RS

\(\left | \vec{QS} \right | > \left | \vec{QR} – \vec{QP} \right | (QP = RS)\)

\(| \vec{a} + \vec{b} |> | | \vec{a} | – | \vec{b} | |\) ...............(iv)

If the two vectors \(\vec{a}\) and \(\vec{b}\) act along a straight line in the same direction, then:

\(\vec{a} + \vec{b} | = | | \vec{a} | – | \vec{b} | | \) ..........(v)

**Combine equation (iv) and (v):**

**\(| \vec{a} + \vec{b} | \geq | | \vec{a} | – | \vec{b} | \)**

**(c)** Let two vectors \(\vec{a}\) and \(\vec{b}\) be represented by the adjacent sides of a** **parallelogram PQRS, as given in the figure.

Here,

**\(\left | \vec{PQ} \right | = \left | \vec{SR} \right | = \left | \vec{b} \right |\)—– (i)**

**\(\left | \vec{PS} \right | = \left | \vec{a} \right |\)—– (ii)**

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in \(\Delta PSR,\)

PR < PS + SR

\(\left | \vec{a} – \vec{b} \right | < \left | \vec{a} \right | + \left |- \vec{b} \right | \left | \vec{a} – \vec{b} \right | < \left | \vec{a} \right | + \left | \vec{b} \right |\) **—– (iii)**

If the two vectors act along a straight line in the opposite direction, then:

\(\left | \vec{a} – \vec{b} \right | = \left | \vec{a} \right | + \left | \vec{b} \right |\) **—– (iv)**

**Combine (iii) and (iv),**

\(\left | \vec{a} – \vec{b} \right | \leq \left | \vec{a} \right | + \left | \vec{b} \right |\)

**(d)** Let two vectors \(\vec{a}\;and \;\vec{b}\)be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

PR + SR > PS **—– (i)**

PR > PS – SR** —– (ii)**

\(\left | \vec{a} – \vec{b} \right | > \left | \vec{a} \right | – \left | \vec{b} \right |\) **—– (iii)**

The quantity on the left hand side is always positive and that on the right hand side can be positive or negative. We take modulus on both the sides to make both quantities positive:

**\(\left |\left | \vec{a} – \vec{b} \right | \right | > \left |\left | \vec{a} \right | – \left | \vec{b} \right | \right | \left | \vec{a} – \vec{b} \right | > \left |\left | \vec{a} \right | – \left | \vec{b} \right | \right |\)—– (iv)**

If the two vectors act along a straight line in the opposite direction, then:

**\(\left | \vec{a} – \vec{b} \right | = \left |\left | \vec{a} \right | – \left | \vec{b} \right | \right |\)—– (v)**

**Combine (iv) and (v):**

\(\left | \vec{a} – \vec{b} \right | \geq \left |\left | \vec{a} \right | – \left | \vec{b} \right | \right |\)

Answered by Abhisek | 1 year ago**(a) **Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

\(\theta (t)=tan^{-1}(\dfrac{v_{0y-gt}}{v_{ox}})\)

**(b)** Shows that the projection angle θ_{0} for a projectile launched from the origin is given by

\(\theta_{0}=tan^{-1}(\dfrac{4h_{m}}{R})\)

where the symbols have their usual meaning

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of a radius of 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

A fighter plane flying horizontally at an altitude of 1.5 km with a speed of 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s^{-1} to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit? (Take g = 10 m s^{-2} ).

A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

Can we associate a vector with

**(i)** a sphere

**(ii)** the length of a wire bent into a loop

**(iii)** a plane area

Clarify for the same.