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(B) Let the direction cosines of the required line be $(l, m, n)$.Since the line is equally inclined to the three mutually perpendicular lines with di

Vedclass · Accepted Answer

$\frac{l_1 + l_2 + l_3}{\sqrt{3}}, \frac{m_1 + m_2 + m_3}{\sqrt{3}}, \frac{n_1 + n_2 + n_3}{\sqrt{3}}$

Vedclass · Answer

(B) Let the direction cosines of the required line be $(l, m, n)$.Since the line is equally inclined to the three mutually perpendicular lines with direction cosines $(l_1, m_1, n_1)$,$(l_2, m_2, n_2)$,and $(l_3, m_3, n_3)$,the cosine of the angle $	heta$ between the required line and each of these lines is the same.Let $\cos 	heta = k$.Then,$l l_1 + m m_1 + n n_1 = k$,$l l_2 + m m_2 + n n_2 = k$,and $l l_3 + m m_3 + n n_3 = k$.Squaring and adding these equations,we get:$(l l_1 + m m_1 + n n_1)^2 + (l l_2 + m m_2 + n n_2)^2 + (l l_3 + m m_3 + n n_3)^2 = 3k^2$.Using the property of orthogonal matrices,for any vector $(l, m, n)$,$\sum (l l_i + m m_i + n n_i)^2 = l^2 + m^2 + n^2 = 1$.Thus,$3k^2 = 1 \implies k = \frac{1}{\sqrt{3}}$.Now,consider the vector $\vec{v} = (l_1+l_2+l_3, m_1+m_2+m_3, n_1+n_2+n_3)$.The magnitude of this vector is $\sqrt{(l_1+l_2+l_3)^2 + (m_1+m_2+m_3)^2 + (n_1+n_2+n_3)^2}$.Since the lines are mutually perpendicular,the sum of squares simplifies to $1+1+1 = 3$.Thus,the unit vector along this direction is $\left( \frac{l_1+l_2+l_3}{\sqrt{3}}, \frac{m_1+m_2+m_3}{\sqrt{3}}, \frac{n_1+n_2+n_3}{\sqrt{3}} ight)$.

The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as $(l_1, m_1, n_1)$,$(l_2, m_2, n_2)$ and $(l_3, m_3, n_3)$ are:

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