Question

The resultant of two vectors at an angle 150⁰ is 10 units and is perpendicular to one vector. Find the magnitude of the smaller vector.
(A) $ 10 units $
(B) $ 10 \sqrt{3} units $
(C) $ 10 \sqrt{2} units $
(C) $ 5 \sqrt{3} units $

Answer 1

Hint
Here we will use the dot vector formula as well as cross vector formula. The Cross Product gives a vector answer, and is sometimes called the vector product.

Complete step by step answer
Let the two vector is $ \mathop A\limits^ \to $ and $ \mathop B\limits^ \to $
According to the question the angle between two vectors is 150⁰ and the resultant vector is perpendicular to one vector. Let the vector is $ \mathop A\limits^ \to $ . And the resultant vector is $ \mathop R\limits^ \to $
Now, the angle between resultant vector and $ \mathop B\limits^ \to $ is $(150-90)=60⁰$
Say,According to the vector rule
$\Rightarrow \left| {\mathop B\limits^ \to } \right|\sin 60^\circ = \left| {\mathop A\limits^ \to } \right| $
Now,
$\Rightarrow \left| {\mathop B\limits^ \to } \right|\cos 60^\circ = \left| {\mathop R\limits^ \to } \right| = 10 $
$\Rightarrow \dfrac{{\left| {\mathop B\limits^ \to } \right|}}{2} = 10 $
$\Rightarrow \left| {\mathop B\limits^ \to } \right| = 20 $
Now,
$\Rightarrow \left| {\mathop A\limits^ \to } \right| = \left| {\mathop B\limits^ \to } \right|\sin 60^\circ = 20 \times \dfrac{{\sqrt 3 }}{2} $
$ \therefore \left| {\mathop A\limits^ \to } \right| = 10\sqrt 3 $ unit
The magnitude of the smaller vector is $ \therefore \left| {\mathop A\limits^ \to } \right| = 10\sqrt 3 $ unit.
Option (B) is correct.

Note
A vector is an object that has both a magnitude and a direction. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.