Question

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the points $(2,3,4)$ and $(1,1,4)$ respectively from the plane $3x-6y+2z+11=0$, then $p_1$, $p_2$ are the roots of the equation

• A. $p^2-23p+7=0$
• B. $7p^2-23p+16=0$
• C. $p^2-17p+16=0$
• D. $p^2-16p+7=0$

Answer 1

Answer: (B)

$7p^2-23p+16=0$

we have

$p_1=\frac{3\times 2-6\times 3+2\times 4+11}{\sqrt{3^2+(-6{)}^2+(2{)}^2}}=\frac{7}{7}=1$

$p_2=\frac{3\times 1-6\times 1+2\times 4+11}{\sqrt{3^2+(-6{)}^2+(2{)}^2}}=\frac{16}{7}$

So that $p_1,p_2$ are the roots of the equation

$p^2-(1+\frac{16}{7})p+\frac{16}{7}=0$

$\Rightarrow 7p^2-23p+16=0$