Consider the line L given by the equation $\frac{x - 3}{2} = \frac{y - 1}{1} = \frac{z - 2}{1} .$ Let Q be the mirror image of the point (2, 3, -1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is one the plane P?

Option 1 - (1, 2, 2)

Option 2 - (-1, 1, 2)

Option 3 - (1, 1, 1)

Option 4 - (1, 1, 2)

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Vishal Baghel | Contributor-Level 10

7 months ago

Correct Option - 1

Detailed Solution:

Equation of plane $⊥^{r}$ to line $\frac{x - 3}{2} = \frac{x - 1}{1} = \frac{x - 2}{1}$ and passes through the point

(2, 3, -1) is 2(x – 2) + 1(y – 3) + 1 (z + 1) = 0

=> 2x + y + z – 6 = 0 ...........(i)

Hence point (1, 2, 2) satisfies equation (i)

$z = \frac{1}{1 - c o s θ + 2 i s i n θ} = \frac{1 - c o s θ - 2 i s i n θ}{{(1 - c o s θ)}^{2} - {(2 i s i n θ)}^{2}} = \frac{2 s i n^{2} \frac{θ}{2} - 2 i s i n θ}{{(1 - c o s θ)}^{2} + 4 s i n^{2} θ}$

$∴ R e (z) = \frac{2 s i n^{2} \frac{θ}{2}}{4 s i n^{2} \frac{θ}{2} (s i n^{2} \frac{θ}{2} + 4 c o s^{2} \frac{θ}{2})} = \frac{1}{2 (1 + 3 c o s^{2} \frac{θ}{2})} = \frac{1}{5} \Rightarrow c o s^{2} \frac{θ}{2} = \frac{1}{2}$

$\Rightarrow θ = \frac{π}{2} ∴ \int_{0}^{\frac{π}{2}} s i n x d x = 1$

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