Question

If the d.r.s of $OA,OB$ are $1,1, - 1;1, - 2,1$ then the d.c.s of the normal to the plane $AOB$ is:
A) $\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}$
B) $\dfrac{2}{{\sqrt {29} }},\dfrac{3}{{\sqrt {29} }},\dfrac{4}{{\sqrt {29} }}$
C) $\dfrac{2}{{\sqrt {14} }},\dfrac{{ - 2}}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}$
D) $\dfrac{1}{{14}},\dfrac{2}{{14}},\dfrac{3}{{14}}$

Answer 1

Hint:
We will use the normality condition of the direction ratios of the given vectors to the given plane. Then we will find the general expression for the obtained plane. We will solve the obtained equations to find the unknown variables and obtain the normal to the plane as required.

Complete step by step solution:
Let the general equation of the plane be $ax + by + cz + d = 0$.
It is given that the direction ratios of $OA,OB$ are $1,1, - 1;1, - 2,1$.
We will assume that the point $O$ corresponds to $\left( {0,0,0} \right)$.
Then the point $A$ will correspond to the point $\left( {1,1, - 1} \right)$.
Similarly, the point $B$ corresponds to the point $\left( {1, - 2,1} \right)$.
This will clearly apply that $d = 0$.
Now from the vector $OA$ we can write the first equation by substituting $\left( {x,y,z} \right) \equiv \left( {1,1, - 1} \right)$.
$a + b - c = 0$ … (1)
Similarly, we will put $\left( {x,y,z} \right) \equiv \left( {1, - 2,1} \right)$.
$a - 2b + c = 0$ … (2)
From equation (1) we get $c = a + b$.
Now substitute this in equation (2) to obtain the following:
$2a - b = 0 \Rightarrow 2a = b$
Now we will express each and every unknown in terms of the variable $b$ as shown below.
$a = \dfrac{1}{2}b,b = b$ and $c = \dfrac{3}{2}b$
Now we have obtained the values of all the unknowns from the equation of the plane.
Therefore, the equation of the plane is $\dfrac{1}{2}bx + by + \dfrac{3}{2}bz = 0$.
We can take out the term $b$ common and divide both sides by $b$ to obtain the equation of the plane as follows:
$\dfrac{1}{2}x + y + \dfrac{3}{2}z = 0$
Therefore, the normal is given by multiplying all the direction ratios of the obtained plane by $2$ .
Thus, the normal is $\vec n = i + 2j + 3k$.
Therefore, the normal unit vector is given by dividing the above equation of the vector by its magnitude.
Thus, the normal unit vector is $\hat n = \dfrac{1}{{\sqrt {1 + 4 + 9} }}i + \dfrac{2}{{\sqrt {1 + 4 + 9} }}j + \dfrac{3}{{\sqrt {1 + 4 + 9} }}k$ .
Which is same as $\hat n = \dfrac{1}{{\sqrt {14} }}i + \dfrac{2}{{\sqrt {14} }}j + \dfrac{3}{{\sqrt {14} }}k$.
This implies, the direction cosines of the plane $AOB$ are $\left\langle {\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}} \right\rangle $.

Hence, the correct option is A.

Note:
Here first we used the values of the direction ratios of the given vectors to find the direction ratio of the normal to the required plane. Now the direction ratios of the plane are the same as the direction ratios of the unit normal vector corresponding to the plane therefore, we just have used the correct unit normal vector to reach the final answer.