Question

Lines $OA, OB$ are drawn from $O$ with direction cosines proportional to $(1, - 2, - 1),(3, - 2,3)$ . Find the direction cosines of the normal to the plane $AOB$
A) \[\left\langle { \pm \dfrac{4}{{\sqrt {29} }}, \pm \dfrac{3}{{\sqrt {29} }},\left. { \pm \dfrac{2}{{\sqrt {29} }}} \right\rangle } \right.\]
B) \[\left\langle { \pm \dfrac{2}{{\sqrt {29} }}, \pm \dfrac{3}{{\sqrt {29} }},\left. { \pm \dfrac{{ - 2}}{{\sqrt {29} }}} \right\rangle } \right.\]
C) \[\left\langle { \pm \dfrac{8}{{\sqrt {29} }}, \pm \dfrac{6}{{\sqrt {29} }},\left. { \pm \dfrac{{ - 2}}{{\sqrt {29} }}} \right\rangle } \right.\]
D) \[\left\langle { \pm \dfrac{8}{{\sqrt {29} }}, \pm \dfrac{3}{{\sqrt {29} }},\left. { \pm \dfrac{{ - 2}}{{\sqrt {29} }}} \right\rangle } \right.\]

Answer 1

Hint:
For solving this particular question, we have to consider the equation of the plane, then according to information we have to form a certain equation. Find the relationship then substitute the values in the equation of the plane , lastly find the direction cosines. In geometry, the direction cosines (or directional cosines) of a vector are the cosines of the angles between the vector and also the three coordinate axes. Equivalently, they're the contributions of every component of the idea to a unit vector in this direction. Direction cosines are a similar extension of the same old notion of slope to higher dimensions.

Complete step by step solution:
Let the plane be $ax + by + cz + d = 0..........(1)$
It is given that plane passes through $O(0,0,0)$,
Therefore, we can say $d = 0$,
It is also given that plane passes through $A(1, - 2, - 1)$,
Substitute the given points in $(1)$ we have ,
$a - 2b - c = 0.............(2)$
It is also given that plane passes through $B(3, - 2,3)$,
Substitute the given points in $(1)$ we have ,
$3a - 2b + 3c = 0.............(3)$
From $(2)$ we have \[c = a - 2b\] , now substitute this in equation $(3)$ , we will get ,
$6a = 8b$
$ \Rightarrow a = \dfrac{4}{3}b,b = b,c = - \dfrac{{2b}}{3}$
Now, substitute these values in the equation $(1)$ , we will get ,
$
   \Rightarrow \dfrac{4}{3}bx + by - \dfrac{2}{3}bz = 0 \\
   \Rightarrow b\left( {\dfrac{4}{3}x + y - \dfrac{2}{3}z} \right) = 0 \\
   \Rightarrow \dfrac{4}{3}x + y - \dfrac{2}{3}z = 0 \\
   \Rightarrow 4x + 3y - 2z = 0 \\
 $
Therefore, normal is given as $ \pm \left( {\mathop n\limits^ \to = 4\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge - 2\mathop k\limits^ \wedge } \right)$
$\mathop n\limits^ \wedge = \pm \left( {\dfrac{4}{{\sqrt {29} }}\mathop i\limits^ \wedge + \dfrac{3}{{\sqrt {29} }}\mathop j\limits^ \wedge - \dfrac{2}{{\sqrt {29} }}\mathop k\limits^ \wedge } \right)$

Therefore, option $A$ \[\left\langle { \pm \dfrac{4}{{\sqrt {29} }}, \pm \dfrac{3}{{\sqrt {29} }},\left. { \pm \dfrac{2}{{\sqrt {29} }}} \right\rangle } \right.\] is the correct option.

Note:
Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. From the normal equation we can easily get the cosines directions as by dividing the normal equation by its magnitude.