Question

If l1,m1,n1,l2,m2,n2,l3,m3,n3 are direction cosines of three mutually perpendicular lines,prove that the line whose direction cosines are proportional to l1+l2+l3,m1+m2+m3,n1+n2+n3 make equal angles with them.

Answer 1

Answer:

L1,M1,N1 and L2,M2,N2 are two mutually perpendicular lines. So, it can be written as $L1^{2} + M1^{2} + N1^{2} = 0$

Similar concept has been considered with l1+l2+l3,m1+m2+m3,n1+n2+n3 to form equal angles.

The direction cosines would be M1N2-M2N1, N1L2-N2L1, L1M2-L2M1.

The whole steps of the equations have been well explained in the below attachments.

Answer 2

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

We can see that the first 3 lines are mutually perpendicular, dot products of pairs of lines are zero.

|1*|2+m1*m2+n1*n2 = 0

|2*| 3+ m2*m3+n2*n3 = 0

|1*|3+m1*m3+n1*n3=0

So if we use the above equations and another dot product

We know that the cosine of the angle between the fourth vector and any of the first three vectors is the same.

So it will make the equal angles with them.

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