Question

if l1,m1,n1 and ĺ2, m2, n2 are the direction cosines of two mutually perpendicular to both of them are m1n2-n1m2, n1l2-l1n2, l1m2-m1l2.

Answer 1

hii friend......here is your answer......

Let <l1, m1, n1> and <l2, m2, n2> be the d.cs. of two mutually perpendicular straight lines.
Then l1l2+m1m2+n1n2=0 ................ (1)
Let <l, m, n> be the d.cs.of the straight lune perpendicular to the above lines.
Then ll1+mm1+nn1=0............................ (2)
ll2+mm2+nn2=0 ...................... (3)

solving (2) and (3) for l, m, n we get
l/(m1n2-m2n1)=m/(n1l2-l1n2)=n/(l1m2-m1l2)

now,

(m1n2-n1m2)^2+(n1l2-l1n2)^2+(l1m2-m1l2)^2
=(l1^2+m1^2+n1^2)(l2^2+m2^2+n2^2)
=(l1l2+m1m2+n1n2)^2
=1.1-0=1


hence m1n2-n2m2, n1l2-l1n2, l1m2-m1l2 are d.cs. of the straight line.

__________________________
I think it will help you then you mark me brainlist please please please please. ..
★★★★★★★

Answer 2

Thank you for asking this question. Here is your answer:

Vectors along the given lines are  

(a↑) = ( l₁, m₁, n₁ ) and (b↑) = ( l₂, m₂, n₂ ).  

A line which is perpendicular to both (a↑) and (b↑)  

is along the cross product (a↑) x (b↑) =  

| i .. j .... k .|  

| l₁ ..m₁..n₁ | = ( m₁n₂ - m₂n₁ ) i - ( l₁n₂ - l₂n₁ ) j + ( l₁m₂ - l₂m₁ ) k ......... (1)  

| l₂ ..m₂..n₂ |  

Coefficients of i, j, k in (1) are the required d.C.s.