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  2. Stuck here, help me understand: - Kinematics - JEE Main-2
# Engineering

Let \left | \overrightarrow{A_{1}} \right |=3,\left | \overrightarrow{A_{2}} \right |=5 and \left | \overrightarrow{A_{1}} +\overrightarrow{A_{2}} \right |=5 the value of \left | 2\overrightarrow{A_{1}} +3\overrightarrow{A_{2}} \right |\cdot \left | 3\overrightarrow{A_{1}} -2\overrightarrow{A_{2}} \right | is

  • Option 1)

    -106.5

  • Option 2)

    -99.5

  • Option 3)

    -112.5

  • Option 4)

    -118.5

 

Answers (1)
S solutionqc

 

Scalar , Dot or Inner Product -

Scalar product of two vector \vec{A} & \vec{B} written as \vec{A} \cdot \vec{B} is a scalar quantity given by the product of magnitude of \vec{A} & \vec{B} and the cosine of smaller angle between them.

\vec{A}\cdot \vec{B}= A\, B\cdot \cos \Theta

- wherein

showing representation of scalar products of vectors.

 

 

\left | \overrightarrow{A_{1}} \right |=3,\left | \overrightarrow{A_{2}} \right |=5, \left | \overrightarrow{A_{1}} + \overrightarrow{A_{2}} \right |=5

\left ( A_{1}+A_{2} \right ).\left ( A_{1}+A_{2} \right )=\left | \overrightarrow{A_{1}} + \overrightarrow{A_{2}} \right |^{2}

\left | \overrightarrow{A_{1}} \right |^{2}+ \left |\overrightarrow{A_{2}} \right |^{2} +2\left ( A_{1}.A_{2} \right )=\left ( 5 \right )^{2}

3^{2}+ 5^{2} +2\left ( A_{1}.A_{2} \right )=25

\left ( A_{1}.A_{2} \right )=\frac{-9}{2}

\left ( 2\overrightarrow{A_{1}}+3\overrightarrow{A_{2}} \right ).\left ( 3\overrightarrow{A_{1}}-2\overrightarrow{A_{2}} \right )

=6A_{1}^{2}-6A_{2}^{2}-4A_{1}A_{2}+9A_{1}.A_{2}

=6(9)-6(25)+5(\frac{-9}{2})=-118.5

 

 


Option 1)

-106.5

Option 2)

-99.5

Option 3)

-112.5

Option 4)

-118.5

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