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Explain solution RD sharma class 12 chapter 23 scalar or dot product exercise Fill in the blanks question 15

Answers (1)

4

Hint:

               \left |a \right |=?

Given:

If a,b non-zero vector of same magnitude. Such that angle between \vec{a} and \vec{b} are \frac{2\pi}{3} and a.b=-8 then  \left | a \right |

Solution:

Let     

\left | a \right |=m\\ \vec{a}.\vec{b}=-8\\ \left |\vec{a} \right |.\left |\vec{b} \right |.\cos \theta =-8\\

m^{2}\left ( -\frac{1}{2} \right ) =-8\\,since a and b have same magnitude

        m^{2}=16

                         m=4,m cannot be negative

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