Q.26. The equation of the plane through (-1,1,2) , whose normal makes equal acute angles with co-ordinate axes is
A. x+y+z-3 = 0
B. x+y+z-2 =0
C. x+y-z-2 = 0
D. x-y+z-3 = 0
Answer :- B. x+y+z-2 =0
Explanation : – Note that (-1,1,2) is satisfied by only option (B)
Alternate method:
Let A = (-1, 1, 2)
\therefore \overrightarrow{a} = -\hat{i} + \hat{j} + 2\hat{k} \overrightarrow{n} = \hat{i} + \hat{j} + \hat{k}\therefore equation of the plane is \vec{r} \cdot \overrightarrow{n}=\vec{a} \cdot \vec{n}
\Rightarrow \vec{r} \cdot (\hat{i}+\hat{j}+\hat{k})=(-\hat{i}+\hat{j}+2\hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k}) \Rightarrow \vec{r} \cdot (\hat{i}+\hat{j}+\hat{k})=2 \Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k})=2 \Rightarrow x+y+z-2=0