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 d² is equal to :

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

Answer 1

The Correct Option is C

Let <a, b, c> be direction ratios of plane containing lines
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)
and
\(\frac{x}{3}=\frac{y}{7}=\frac{z}{8}\).
∴ 2a + 3b + 5c = 0 …(i)
and 3a + 7b + 8c = 0 …(ii)
from eq. (i) and (ii)
\(\frac{a}{24-35}\)=\(\frac{b}{15-16}\)=\(\frac{c}{14-9}\)
∴ D.R^s. of plane are < 11, 1, –5>
Let D.R^S of plane P be <a₁, b₁, c₁> then.
11a₁ + b₁ – 5c₁ = 0 …(iii)
and 9a₁ – b₁ – 5c₁ = 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 + 2z – 21 = 0
Distance from point (2, –5, 11) is
d=\(\frac{|2+5+22−2|}{\sqrt6}\)
∴d²=\(\frac{32}{3}\)