Question

The direction ratios of two lines AB, AC are $1,-1,-1\text{ and 2,-1,1}\text{.}$The direction ratios of the normal to the plane ABC are
$\begin{aligned} & A.\text{ 2,3,-1} \\\ & \text{B}\text{. 2,2,1} \\\ & \text{C}\text{. 3,2,-1} \\\ & \text{D}\text{. -1,2,3} \\\ \end{aligned}$

Answer 1

Hint: Any line whose direction ratio is $a,b,c$can be written in vector form as $a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}$. So first of all, write the given lines in vector form as $\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k}$and $\overrightarrow{2i}-\overrightarrow{j}+\overrightarrow{k}$.find the cross product of the vectors. This gives a vector which is perpendicular to both vectors, in other words it is perpendicular to the plane containing line AB and AC. The components of these vectors along the x,y and z axis is the direction ratio of the normal to the plane ABC.

Complete step-by-step answer:
As it is given from question direction ratio of line AB is 1, -1, -1
So, we can write
$\overrightarrow{AB}=\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{z}$, where $\overrightarrow{i},\overrightarrow{j},\overrightarrow{k}$are the unit vectors along x, y, z axis respectively.
Similarly, we can write,
$\overrightarrow{AC}=2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}$
Now any vector which is perpendicular $\overrightarrow{AB}=\overrightarrow{i}-\overrightarrow{j}-\overrightarrow{z}$and$\overrightarrow{AC}=2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}$both is the cross product of two vectors.
Now
$\overrightarrow{AB}\times \overrightarrow{AC}=$$$\begin{aligned}
& \overrightarrow{AB}\times \overrightarrow{AC}=\left( \overrightarrow{i}-\overrightarrow{j}-\overrightarrow{k} \right)\times \left( 2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k} \right) \\
& \Rightarrow \overrightarrow{AB}\times \overrightarrow{AC}=\overrightarrow{i}\times \left( 2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k} \right)-\overrightarrow{j}\times \left( 2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k} \right)-\overrightarrow{k}\times \left( 2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k} \right) \\
\end{aligned}$$

So, we can write

& \overrightarrow{AB}\times \overrightarrow{AC}=\overrightarrow{i}\times \left( 2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k} \right)-\overrightarrow{j}\times \left( 2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k} \right)-\overrightarrow{k}\times \left( 2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k} \right) \\\ & \overrightarrow{AB}\times \overrightarrow{AC}=2\overrightarrow{i}\times \overrightarrow{i}-\overrightarrow{i}\times \overrightarrow{j}+\overrightarrow{i}\times \overrightarrow{k}-2\overrightarrow{j}\times \overrightarrow{i}+\overrightarrow{j}\times \overrightarrow{j}-\overrightarrow{j}\times \overrightarrow{k}-2\overrightarrow{k}\times \overrightarrow{i}+\overrightarrow{k}\times \overrightarrow{j}-\overrightarrow{k}\times \overrightarrow{k} \\\ \end{aligned}$$ As we know the properties of cross product of unit vectors as $$\begin{aligned} & \overrightarrow{i}\times \overrightarrow{j}=\overrightarrow{k},\overrightarrow{k}\times \overrightarrow{i}=\overrightarrow{j},\overrightarrow{j}\times \overrightarrow{k}=\overrightarrow{i} \\\ & \overrightarrow{i}\times \overrightarrow{i}=\overrightarrow{0},\overrightarrow{k}\times k=\overrightarrow{0},\overrightarrow{j}\times \overrightarrow{j}=\overrightarrow{0} \\\ & \overrightarrow{j}\times \overrightarrow{i}=-\overrightarrow{k},\overrightarrow{i}\times \overrightarrow{k}=-\overrightarrow{j},\overrightarrow{k}\times \overrightarrow{j}=-\overrightarrow{i} \\\ \end{aligned}$$ So, using the above properties we can write $$\begin{aligned} & \overrightarrow{AB}\times \overrightarrow{AC}=\overrightarrow{0}-\overrightarrow{k}-\overrightarrow{j}+2\overrightarrow{k}+\overrightarrow{0}-\overrightarrow{i}-2\overrightarrow{j}-\overrightarrow{i}-\overrightarrow{0} \\\ & \Rightarrow \overrightarrow{AB}\times \overrightarrow{AC}=+\overrightarrow{k}-3\overrightarrow{j}-2\overrightarrow{i} \\\ & \Rightarrow \overrightarrow{AB}\times \overrightarrow{AC}=-2\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k} \\\ \end{aligned}$$ Hence, we can write $$-2\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k}$$is a vector which is perpendicular to $$\overrightarrow{AB}\text{ and}\overrightarrow{AC}$$that is the line AB and AC, so we can say that $$-2\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k}$$is the vector which is perpendicular to the plane ABC. So the direction ratios of the normal to the plane ABC are the components of the vector along the x, y, z axis. $$-2\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k}$$is perpendicular to plane ABC and towards the viewer. $$+2\overrightarrow{i}+3\overrightarrow{j}-\overrightarrow{k}$$is perpendicular to plane ABC and away from the viewer that is below the plane ABC. So as per option we can say that 2,3, -1 is the required direction ratio. so, option A is correct. NOTE. When we find the cross product of two vectors, we get a vector which is perpendicular to both vectors.so it is toward the viewer or away from the viewer. If the order of the cross product of the vector reversed then direction of perpendicular vector also reversed. If A and B are two vectors, we have $\overrightarrow{A}\times \overrightarrow{B}=-\overrightarrow{B}\times \overrightarrow{A}$so, in this question $\overrightarrow{AC}\times \overrightarrow{AB}=-\overrightarrow{AB}\times \overrightarrow{AC}$