Question

The distance of the point P (3,8,2) from the line $\frac{1}{2}(x-1)=\frac{1}{4}(y-3)=\frac{1}{3}(z-2)$ measured parallel to the plane $3x+2y-2z+15=0is$

• A. 8 units
• B. 5 units
• C. 6 units
• D. 7 units

Answer 1

The correct option is D

7 units

Let a general point on line be $A(2\lambda +1,4\lambda +3,3\lambda +2)$.
Let this point lie at the same distance as the point P(3, 8, 2) from the plane 3x+ 2y- 2z+ 15= 0

Therefore, $\frac{3\times 3+2\times 8-2\times 2+15}{\sqrt{17}}=\frac{3(2\lambda +1)+2(4\lambda +3)-2(3\lambda +2)+15}{\sqrt{17}}\Rightarrow 36=8\lambda +20\Rightarrow \lambda =2$
Therefore, A is (5, 11, 8).
$PA=\sqrt{(5-3{)}^2+(11-8{)}^2+(8-2{)}^2}=\sqrt{4+9+36}=7$