10) Let λ 0.60 i .80 j and . 1 → = ˆ + 0 ˆ λ 0.50 i .866 j 2 → = ˆ + 0 ˆ 1. Show that both the vectors are unit vectors. 2. Is the sum of these two unit vectors also a unit vector? If not, find a unit vector along the sum of λ and . 1 → λ2

Answer:2.  [tex]0.55099\ \hat i+0.8345\ \hat j.[/tex]Step-by-step explanation:Given vectors are:[tex]\vec \lambda_1=0.60\ \hat i+0.80\ \hat j.[/tex][tex]\vec \lambda_2 = 0.50\ \hat i+0.866\ \hat j.[/tex]where, [tex]\hat i,\ \hat j[/tex] are the unit vectors along positive x and y axes respectively.(1):A vector is called unit vector if its magnitude (length) is 1 and the magnitude of a vector [tex]\vec a = a_x\ \hat i+a_y\ \hat j[/tex] is given by[tex]a=\sqrt{a_x^2+a_y^2}.[/tex]Therefore, The magnitude of [tex]\vec \lambda_1[/tex] = [tex]\sqrt{0.60^2+0.80^2}=1.00.[/tex]The magnitude of [tex]\vec \lambda_2[/tex] =[tex]\sqrt{0.50^2+0.866^2}=0.999978\approx 1.00.[/tex]Thus, both the vectors are unit vectors.(2):The sum of these vectors is given by[tex]\vec \lambda_1+\vec \lambda_2=(0.60\hat i+0.80\hat j)+(0.50\hat i+0.866\hat j)\\=(0.60+0.50)\hat i+(0.80+0.866)\hat j\\=1.10\ \hat j+1.666\ \hat j.[/tex]The magnitude of the sum of these vectors = [tex]\sqrt{1.10^2+1.666^2}=1.9964.[/tex]Thus, it is not a unit vector.The unit vector along the direction of sum of these two vectors is given by[tex]\hat \lambda =\dfrac{\vec \lambda_1+\vec \lambda_2}{|\vec \lambda_1+\vec \lambda_2|}=\dfrac{1.10\ \hat j+1.666\ \hat j}{1.9964}=0.55099\ \hat i+0.8345\ \hat j.[/tex]