Question:
Given two vectors $\vec{ A }=-\hat{ i }+2 \hat{ j }-3 \hat{ k }$ and $\vec{ B }=4 \hat{ i }-2 \hat{ j }+6 \hat{ k }$. The angle made by $(\vec{ A }+\vec{ B })$ with $x$-axis is
Given two vectors $\vec{ A }=-\hat{ i }+2 \hat{ j }-3 \hat{ k }$ and $\vec{ B }=4 \hat{ i }-2 \hat{ j }+6 \hat{ k }$. The angle made by $(\vec{ A }+\vec{ B })$ with $x$-axis is
Updated On: Jul 5, 2022
- $30^{\circ}$
- $45^{\circ}$
- $60^{\circ}$
- $90^{\circ}$
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The Correct Option is B
Solution and Explanation
Given, $\vec{ A }=-\hat{ i }+2 \hat{ j }-3 \hat{ k } $
$\vec{ B }=4 \hat{ i }-2 \hat{ j }+6 \hat{ k }$
Let angle made by $(\vec{ A }+\vec{ B })$ with $x-$ axis is $\theta$.
$\cos \theta=\frac{(\vec{ A }+\vec{ B }) \cdot \hat{ i }}{|\vec{ A }+\vec{ B }| \cdot|\hat{ i }|} $
$=\frac{(3 \hat{ i }+3 \hat{ k }) \cdot i}{\sqrt{9+9 \cdot 1}}$
$=\frac{3}{3 \sqrt{2}}=\frac{1}{\sqrt{2}} $
$\therefore \theta=45^{\circ}$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration