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.
Solution
Let the two vectors be \[\vec{a}\] and \[\vec{b}\]
Now,
\[\left| \vec{a} \right| = 3 \text { and } \left| \vec{b} \right| = 4\]
\[\sqrt{\vec{a}^2 + \vec{b}^2 + 2 . \vec{a} . \vec{b} \cos \theta} = 1\]
\[ \Rightarrow \sqrt{3^2 + 4^2 + 2 . 3 . 4 \cos \theta} = 1\]
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°.
