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  • It is found that \left | A+B \right |=\left | A \right |. This necessarily implies, (a) B=0 (b) A,B are antiparallel (c) A,B are perpendicular (d) A.B\leq 0

It is found that \left | A+B \right |=\left | A \right |. This necessarily implies,

(a)B=0

(b) A,B are antiparallel

(c) A,B are perpendicular

(d) A.B\leq 0

 

 

Answers (1)

The answer is the option (a) B=0

Explanation :

Squaring both sides and opening the brackets we get,

\left | \vec{A} \right |^{2}+\left |\overrightarrow{B} \right |^{2}+2\left | \vec{A} \right |\left | \overrightarrow{B} \right |\cos \theta =\left | \vec{A} \right |^{2}

\left | \overrightarrow{B} \right |^{2}+2\left | \vec{A} \right |\left | \overrightarrow{B} \right |\cos \theta =0

cos\theta=\frac{-B}{2A}

if B=0

cos\theta=0

 

Posted by

infoexpert24

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