Question

If three mutually perpendicular lines have direction cosines \[\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right),\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right)\] and \[\left( {{l}_{3}},{{m}_{3}},{{n}_{3}} \right)\] , then the line having direction cosines \[{{l}_{1}}+{{l}_{2}}+{{l}_{3}},{{m}_{1}}+{{m}_{2}}+{{m}_{3}}\] and \[{{n}_{1}}+{{n}_{2}}+{{n}_{3}}\] make an angle of ….....with each other.
A. $ 0{}^\circ $
B. $ 30{}^\circ $
C. $ 60{}^\circ $
D. $ 90{}^\circ $

Answer 1

Hint: Using the direction cosines write the vectors of given 3 mutually perpendicular lines. Assume the required line to be some variable. By using the condition of perpendicularity : dot product of2 perpendicular vectors is 0. Find the 3 relations associated to 3 given mutually perpendicular vectors. Now, find the dot product of assumed variable vectors with the given 3 mutually perpendicular vectors. From which you get the angle between them. By dot product rule, we have 2 vectors of x degrees angle between them, we say:
 $ \overrightarrow{a}\cdot \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta $

Complete step-by-step answer:
The given direction cosines of 3 mutually perpendicular vectors are given by the terms: $ \left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)\text{ }\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right)\text{ }\left( {{l}_{3}},{{m}_{3}},{{n}_{3}} \right) $
Let this 3 mutually perpendicular vectors be named a, b, c:
 $ a={{l}_{1}}i+{{m}_{1}}j+{{n}_{1}}k $ ; $ b={{l}_{2}}i+{{m}_{2}}j+{{n}_{2}}k $ ; $ c={{l}_{3}}i+{{m}_{3}}j+{{n}_{3}}k $
The direction cosine of required line for which we need angle:
Let this line be assumed as variable p, (for now)
 $ p=\left( {{l}_{1}}+{{l}_{2}}+{{l}_{3}} \right)\hat{i}+\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)\hat{j}+\left( {{n}_{1}}+{{n}_{2}}+{{n}_{3}} \right)\hat{k} $
By basic knowledge of geometry, if a, b are perpendicular we get : $ \overline{a}\cdot \overline{b}=0 $
As we know a, b, c are mutually perpendicular.
By applying above rule 3 times to 3 possible pairs, we get:
 $ {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0 $ ; $ {{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}}=0 $ ; $ {{l}_{3}}{{l}_{1}}+{{m}_{3}}{{m}_{1}}+{{n}_{3}}{{n}_{1}}=0 $
As l, m, n are direction cosine by general maths, we can say:
 $ {{l}_{1}}^{2}+{{m}_{1}}^{2}+{{n}_{1}}^{2}=1 $ ; $ {{l}_{2}}^{2}+{{m}_{2}}^{2}+{{n}_{2}}^{2}=1 $ ; $ {{l}_{3}}^{2}+{{m}_{3}}^{2}+{{n}_{3}}^{2}=1 $
Case-1: Dot product of a, n
By dot product we know: - $ \overline{a}\cdot \overline{b}=\left| \overline{a} \right|\left| \overline{b} \right|\cos x $
As these are direction cosines, we get $ \left| a \right|=\left| b \right|=\left| c \right|=\left| n \right|=1 $
By applying above if x is angle between a, n then:
 $ \cos x=\left( {{l}_{1}}i+{{m}_{1}}j+{{n}_{1}}k \right)\left( \left( {{l}_{1}}+{{l}_{2}}+{{l}_{3}} \right)i+\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)j+\left( {{n}_{1}}+{{n}_{2}}+{{n}_{3}} \right)k \right) $
By simplifying the dot product above, we get $ \cos x $ to be:
 $ \cos x={{l}_{1}}\left( {{l}_{1}}+{{l}_{2}}+{{l}_{3}} \right)+{{m}_{1}}\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)+{{n}_{1}}\left( {{n}_{1}}+{{n}_{2}}+{{n}_{3}} \right) $
By simplifying more the $ \cos x $ will get converted to:
 $ \cos x={{l}_{1}}^{2}+{{l}_{1}}{{l}_{2}}+{{m}_{1}}^{2}+{{m}_{1}}{{m}_{2}}+{{m}_{1}}{{m}_{3}}+{{n}_{1}}^{2}+{{n}_{1}}{{n}_{2}}+{{n}_{1}}{{n}_{3}} $
By rearranging the terms in above equation, we get it as:
 $ \cos x=\left( {{l}_{1}}^{2}+{{m}_{1}}^{2}+{{n}_{1}}^{2} \right)+\left( {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}} \right)+\left( {{l}_{1}}{{l}_{3}}+{{m}_{1}}{{m}_{3}}+{{n}_{1}}{{n}_{3}} \right) $
By using the values of these expressions as calculated before, we get
 $ \cos x=1+0+0=1 $
By applying $ {{\cos }^{-1}} $ on both sides of the equation, we get $ x=0{}^\circ $ .
Similarly, as only 1, 2, 3 positions change we can say that:
 $ y=z=0{}^\circ $
By above 2 equations, we can say $ x=y=z=0 $ .
Hence, angle made by vector p with a, b, c are equal.
Option (a) is correct.

Note: (1) If you solve p.b, p.c you set the same result.
(2) The grouping of terms to use the equations known is crucial.
(3) Do the dot product carefully use $ i\cdot i=1\text{ }i\cdot j=0\text{ }i\cdot k=0 $ to solve.