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If [latex]\vec{a}=i -j[/latex] , [latex]\vec{b}=i +j+k[/latex] are two vectors, and c is another vector such that [latex]\vec{a}\times \vec{c}+\vec{b}=\vec{0}[/latex] and [latex]\vec{a}. \vec{c}=4[/latex] then [latex]\left | \vec{c}\right |^{2} =[/latex]

a) 9 b) 8 c) 19/2 d) 17/2 Solution: \(\begin{array}{l}\vec{a}\times \vec{c}+\vec{b}=\vec{0}\end{array} \) \(\begin{array}{l}\vec{a}\times \vec{c}=-\vec{b}\end{array} \) \(\begin{array}{l}\vec{a\times }\left ( \vec{a}\times \vec{c}\right... View Article

The sum [latex]sum _{k=1}^{20}(1+2+3+ ……+k)[/latex] is

Solution: \(\begin{array}{l}\sum_{k=1}^{20}\frac{k(k+1)}{2}\end{array} \) \(\begin{array}{l}=\frac{1}{2}\sum_{k=1}^{20}k^{2} + k\end{array} \) \(\begin{array}{l}=\frac{1}{2}\left [ \frac{20(21)(41)}{6}+\frac{20(21)}{2} \right ]\end{array} \) =(1/2)[2870+210]=1540