Question

Let P be the plane containing the straight line$\frac{ x−3}{9}$=$\frac{y+4}{-1}$=$\frac{z-7}{-5}$and perpendicular to the plane containing the straight lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$.If d is the distance P from the point (2, –5, 11), then d 2 is equal to :

• A. $\frac{147}{2}$
• B. 96
• C. $\frac{32}{3}$
• D. 54

Answer 1

Answer: (C)

$\frac{32}{3}$

Answer 2

Let <a , b , c > be direction ratios of plane containing lines
$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$
and
$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$.
∴ 2 a + 3 b + 5 c = 0 …(i)
and 3 a + 7 b + 8 c = 0 …(ii)
from eq. (i) and (ii)
$\frac{a}{24-35}$=$\frac{b}{15-16}$=$\frac{c}{14-9}$
∴ D.Rs. of plane are < 11, 1, –5>
Let D.RS of plane P be <a 1, b 1, c 1> then.
11 a 1 + b 1 – 5 c 1 = 0 …(iii)
and 9 a 1 – b 1 – 5 c 1 = 0 …(iv)
From eq. (iii) and (iv) :
$\frac{a_1}{-5-5}$=$\frac{b_1}{-45+55}$=$\frac{c_1}{-11-9}$
∴ D.A5. of plane P are < 1, –1, 2>
Equation plane P is : 1(x – 3) –1(y + 4) +2(z –7) = 0
⇒ x – y + 2 z – 21 = 0
Distance from point (2, –5, 11) is
d=$\frac{|2+5+22−2|}{\sqrt6}$
∴d2=$\frac{32}{3}$