Given:u=(2,6,−7)v=(−1,−1,8)ku+lv=(2,14,11)−−−−−−−−−−(i)ku=k(2,6,−7)=(2k,6k,−7k)lv=l(−1,−1,8)=(−l,−l,8l)ku+lv=(2k−l,6k−l,−7k+8l)−−−(ii)Comparing (i) and (ii),2k−l=26k−l=14−7k+8l=11Hence,2k−l−2=0−−−−−−−−−−−−(iii)6k−l−14=0−−−−−−−−−−−(iv)−7k+8l−11=0−−−−−−−−−−(v)By solving either (iii) and (iv), (iv) and (v),or (iii) and (v) simultaneously, we obtaink=−3,l=−4.Hence, the value of l=−4.
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