If a,b,c,l and m are vectors, prove that
[a b c] `(lxxm)=|(a,b,c),(a*l,b*l,c*l),(a*m,b*m,c*m)|`

If a, b, c, l,m are in A.P, then the value of a-4b + 6c -4l +m is.

If l ,\ m ,\ n are scalars and vec a ,\ vec b ,\ vec c are vectors, prove that |l vec a+m vec b+n vec c|^2=l^2| vec a|^2+m^2| vec b|^2+n^2 | vec c|^2+2\ {l m( vec a . vec b)+m n( vec b . vec c)+n l( vec c . vec a)} . Also deduce that |l vec a+m vec b+n vec c|^2=l^2 |vec a|^2+m^2| vec b|^2+n^2| vec c|^2if\ vec a ,\ vec b ,\ vec c are mutually perpendicular vectors.

Let a, b, c be the side-lengths of a triangle, and l, m,n be the lengths of its medians. Put K=((l+m+n)/(a+b+c)) Then, as a, b, c vary, K can assume every value in the interval

If A=|{:(a,d,l),(b,e,m),(c,f,n):}|" and B = "|{:(l,m,n),(a,b,c),(d,e,f):}| , then

Supose directioncoisnes of two lines are given by u l+vm+wn=0 and al^2+bm^2+cn^2=0 where u,v,w,a,b,c are arbitrary constnts and l,m,n are directioncosines of the lines. For u=v=w=1 if (n_1 n_2)/(l_1 l_2)=((a+b)/(b+c)) then (A) (m_1m_2)/(l_1 l_2)=((b+c))/((c+a)) (B) (m_1m_2)/(l_1 l_2)=((c+a))/((b+c)) (C) (m_1m_2)/(l_1 l_2)=((a+b))/((c+a)) (D) (m_1m_2)/(l_1 l_2)=((c+a))/((a+b))

Supose directioncoisnes of two lines are given by u l+vm+wn=0 and al^2+bm^2+cn^2=0 where u,v,w,a,b,c are arbitrary constnts and l,m,n are directioncosines of the lines. For u=v=w=1 if (n_1 n_2)/(l_1 l_2)=((a+b)/(b+c)) then (A) (m_1m_2)/(l_1 l_2)=((b+c))/((c+a)) (B) (m_1m_2)/(l_1 l_2)=((c+a))/((b+c)) (C) (m_1m_2)/(l_1 l_2)=((a+b))/((c+a)) (D) (m_1m_2)/(l_1 l_2)=((c+a))/((a+b))

If a=x^(m+n)\ y^l,\ b=x^(n+l)\ y^m and c=x^(l+m)y^n , prove that a^(m-n)\ b^(n-l)\ c^(l-m\ )=1

In A B C , A L and C M are the perpendiculars from the vertices A and C to B C and A B respectively. If A L and C M intersect at O , prove that: (i) O M A O L C (ii) (O A)/(O C)=(O M)/(O L)

If a=x^(m+n)y^(1),b-x^(n+l)y^(m) and c=x^(l+m)y^(n), prove that a^(m-n)b^(n-1)c^(1-m)=1

If the vectors vec a, vec b, vec c are non-coplanar and l,m,n are distinct real numbers, then [(l vec a + m vec b + n vec c) (l vec b + m vec c + n vec a) (l vec c + m vec a + n vec b)] = 0, implies (A) lm+mn+nl = 0 (B) l+m+n = 0 (C) l^2 + m^2 + n^2 = 0