Question

Let P be the foot of the perpendicular dropped from the origin O on the line of intersection of the planes $x-2y+3z=5$ and $2x+3y+z+4=0$, then

• A. Equation of the plane perpendicular to the given line and passing through the point (1,1,1) is $11x-5y-7z+1=0$
• B. Equation of acture angle bisector of the two planes is $x+5y-2z+9=0$.
• C. $\overline{OP},\hat{i}-2\hat{j}+3\hat{k}$ and $2\hat{i}+3\hat{j}+\hat{k}$ must be coplanar
• D. The point (1,2,3) lies in acute region of the two planes

Answer 1

Answer: (A), (B), (C), (D)

The point (1,2,3) lies in acute region of the two planes

Since $\overline{OP},\hat{i}-2\hat{j}+3\hat{k}$ and $2\hat{i}+3\hat{j}+\hat{k}$ are perpendicular to the given straight line
$\Rightarrow$ all the three vectors are coplaner
Planes are $-x+2y-3z+5=0....\left(i\right)$
$2x+3y+z+4=0....\left(ii\right)$
$\therefore a_1{a}_2+b_1{b}_2+c_1{c}_2=-2+6-3=1>0$
$\Rightarrow$ Origin lies in obtuse angle
$\therefore$ Equation of the required acute angle bisector is
$\frac{-x+2y-3z+5}{\sqrt{1+4+9}}=-\frac{2x+3y+z+4}{\sqrt{4+5+1}}$
$\Rightarrow -x+2y-3z+5+2x+3y+z+4=0$
$\Rightarrow x+5y-2z+9=0$
Substituting the point (1, 2, 3) is equation of planes (i) and (ii), we ges
$-1+4-9+5<0$ and $2+6+3+4>0$
$\Rightarrow$ the point (1,2,3) lies in acute angle
Let (a, b,c) be the d.r.'s of given line
$\therefore -a+2b-3c=0$
$2a+3b+c=0$
$\therefore \frac{a}{2+9}=\frac{-b}{-1+6}=\frac{c}{-3-4}$
$\Rightarrow \frac{a}{11}=\frac{b}{-5}=\frac{c}{-7}$
$\therefore$ D.r.'s of the line are 11, -5, -7
$\therefore$ Required equation of plane is
$4x-5y-7z+1=0$