Question

6. The velocity of three particles A, B, C are given below, which particle travels at the greatest speed?
Va= 3i-5j+2k Vb=i+2j +3k V c = 5i + 3j +4k​

Answer 1

Given :

The velocity of three particles A, B, C are :

• $\bf{V_a}$ = 3 i + 5 j + 2 k
• $\bf{V_b}$ = i + 2 j + 3 k
• $\bf{V_c}$ = 5 i + 3 j + 4 k

To find :

• Which particle travels with greatest speed

Knowledge required :

• Finding the magnitude of a vector with orthogonal notations

The magnitude of a vector in the form

A = a i + b j + c k

is given by,

$\boxed{\rm{|\bold{A}|=\sqrt{a^2+b^2+c^2}}}$

Solution :

Calculating magnitudes of given velocities of three particles :

__________________________

→ $\bf{V_a}$ = 3 i - 5 j + 2 k

→ | $\bf{V_a}$ | = $\rm{\sqrt{(3)^2+(-5)^2+(2)^2}}$

→ | $\bf{V_a}$ | = $\rm{\sqrt{9 + 25 + 4\;}}$

→ | $\bf{V_a}$ | = $\rm{\sqrt{38}}$

__________________________

→  $\bf{V_b}$ = i + 2 j + 3 k

→ | $\bf{V_b}$ | = $\rm{\sqrt{(1)^2+(2)^2+(3)^2}}$

→ | $\bf{V_b}$ | = $\rm{\sqrt{1+4+9\;}}$

→ | $\bf{V_b}$| = $\rm{\sqrt{14}}$

__________________________

→  $\bf{V_c}$ = 5 i + 3 j + 4 k

→ | $\bf{V_c}$ | = $\rm{\sqrt{(5)^2+(3)^2+(4)^2}}$

→ | $\bf{V_c}$ | = $\rm{\sqrt{25+9+16\;}}$

→ | $\bf{V_c}$ | = $\rm{\sqrt{50}}$

__________________________

As we can see,

Magnitude of V$\bf{{}_c}$ will be greatest therefore,

• Particle C will travel with the greatest speed.