If l1,m1,n1,l2,m2,n2 and l3,m3,n3 are the direction cosines of three mutually perpendicular lines,then prove that the line whose direction cosines are proportional to l1+l2+l3,m1+m2+m3 and n1+n2+n3 makes equal angles with them.
Let →a=l1^i+m1^j+n1^k→b=l2^i+m2^j+n2^k→c=l3^i+m3^j+n3^k→d=(l1+l2+l3)^i+(m1+m2+m3)^j+(n1+n2+n3)^kAlso,let α, β and γ are the angles between →a and →d,→b and →d,→c and →d.∴ cos α=l1(l1+l2+l3)+m1(m1+m2+m3)+n1(n1+n2+n3)=l21+l1l2+l1l3+m21+m1m2+m1m3+n2+n1n2+n1n3=(l21+m21+n21)+(l1l2+l1l3+m1m2+m1m3+n1n2+n1n3)=1+0=1[∵l21+m21+n21=1 and l1⊥l2,l1⊥l3+m1⊥m2,m1⊥m3,n1⊥n2,n1⊥n3]Similarly, cosβ=l2(l1+l2+l3)+m2(m1+m2+m3)+n2(n1+n2+n3)=1+0 and cosγ=1+0⇒ cosα=cosβ=cosγ⇒α=β=γ
So,the line whose direction cosines are proportional to l1+l2+l3,m1+m2+m3,n1+n2+n3 makes equal angles with the three mutually perpendicular lines whose direction cosines are l1m1n1, l2m2n2, l3m3n3, respectively.