Question

Two vectors have magnitudes 2 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.

Answer 1

Let the two vectors be  \[\vec{a}\]  and  \[\vec{b}\]

Now, 

\[\left| \vec{a} \right| = 3 \text { and } \left| \vec{b} \right| = 4\] 

(a) If the resultant vector is 1 unit, then

\[\sqrt{\vec{a}^2 + \vec{b}^2 + 2 . \vec{a} . \vec{b} \cos \theta} = 1\]

\[ \Rightarrow \sqrt{3^2 + 4^2 + 2 . 3 . 4 \cos \theta} = 1\]

Squaring both sides, we get: 

\[25 + 24 \cos \theta = 1\]

\[ \Rightarrow 24 \cos \theta = - 24\]

\[ \Rightarrow \cos \theta = - 1\]

\[ \Rightarrow \theta = 180^\circ\]

Hence, the angle between them is 180°.
(b) If the resultant vector is 5 units, then

\[\sqrt{\vec{a}^2 + \vec{b}^2 + 2 . \vec{a} . \vec{b} \cos \theta} = 5\]

\[ \Rightarrow \sqrt{3^2 + 4^2 + 2 . 3 . 4 \cos \theta} = 5\]

Squaring both sides, we get:

    25 + 24 cos θ = 25
⇒ 24 cos θ = 0
⇒ cos θ = 90°

Hence, the angle between them is 90°.
(c) If the resultant vector is 7 units, then

\[\sqrt{\vec{a}^2 + \vec{b}^2 + 2 . \vec{a} . \vec{b} \cos \theta} = 1\]

\[ \Rightarrow \sqrt{3^2 + 4^2 + 2 . 3 . 4 \cos \theta} = 7\]
Squaring both sides, we get:

 25 + 24 cos θ = 49,
⇒ 24 cos θ = 24
⇒ cos θ = 1
⇒ θ = cos⁻¹ 1 = 0°

Hence, the angle between them is 0°.