Let equation of plane through (1, 2, -4) be
a(x - 1) + b(y - 2) + c(z + 4) = 0
Given lines are
\(\overrightarrow r\) = î + 2ĵ - 4k̂ + λ(2î + 3ĵ + 6k̂)
and \(\overrightarrow r\) = î - 3ĵ + 5k̂+ μ(î + ĵ - k̂)
The cartesian equations of given lines are
\(\frac{x -1}{2} = \frac{y -2}{3} = \frac{z + 4}{6}\ \text{and } \ \frac{x - 1}{1} = \frac{y + 3}{1} = \frac{z - 5}{1}\)
Since, the required plane (j) is parallel in the given lines,
so normal to the plane is perpendicular to the given lines.
∴ 2a + 3b + 6c = 0
and a + b + c = 0
For solving these two equations by cross-multiplication,
we get
\(\frac{a}{-3-6} = \frac{b}{6+2} = \frac{c}{2-3} \implies \frac{a}{-9} = \frac{b}{8} \frac{c}{-1} = \lambda \ \text{(say)}\)
∴ a = - 9λ, b = 8λ, c = - λ
On putting values of a, b and c in Eq. (i), we get
- 9λ(x - 1) + 8λ(y - 2) - λ(z + 4) = 0
Equation of plane in cartesian form is
- 9λ (x - 1) + 8λ (y - 2) - λ(z + 4) = 0
⇒ - 9x + 9 + 8y - 16 - z - 4 = 0
⇒ 9x - 8y + z + 11 = 0
Now, vector from of plane is
\(\overrightarrow r\). (9î - 8ĵ + k̂) = - 11
Also, distance of (9, - 8, - 10) from the above plane
