Question

Let the image of the point $P(2,-1,3)$ in the plane $x+2 y-z=0$ be $Q$ Then the distance of the plane $3 x+2 y+z+29=0$ from the point $Q$ is

• A. $2 \sqrt{14}$
• B. $\frac{22 \sqrt{2}}{7}$
• C. $3 \sqrt{14}$
• D. $\frac{24 \sqrt{2}}{7}$

Answer 1

The Correct Option is C

eq. of line PM 1x−2​=2y+1​=−1z−3​=λ 
any point on line=(λ+2,2λ−1,−λ+3) 
for point ' m ’ (λ+2)+2(2λ−1)−(3−λ)=0 
λ=\(\frac{1}{2}\)

Point m(\(\frac{1}{2}+2,2\times\frac{1}{2},\frac{-1}{2}+3\)) 

=(\(\frac{5}{2},0,\frac{5}{2}\)​) 

For Image Q(α,β,γ) 
​\(\frac{\alpha + 2}{2}=\frac{5}{2},\frac{\beta-1}{2}=0\) 

\(\frac{\gamma +3}{2}=\frac{5}{2}\)

Q:(3,1,2) 
\(d=|\frac{3(3)+2(1)+2+29}{\sqrt{3^2+2^2+1^2}}|\)

\(d=\frac{42}{\sqrt{14}}=3\sqrt{14}\)

Answer 2

Equation of line \( PM \): \[ \frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 3}{-1} = \lambda \] Any point on this line is: \[ (\lambda + 2, 2\lambda - 1, -\lambda + 3) \] Solving for \( \lambda \), we get \( \lambda = \frac{1}{2} \). Thus, the midpoint \( M \) is: \[ \left( \frac{5}{2}, 0, \frac{5}{2} \right) \] For the image \( Q(\alpha, \beta, \gamma) \), we compute: \[ \alpha + 2 = \frac{5}{2}, \quad \beta - 1 = 0, \quad \gamma + 3 = \frac{5}{2} \] \[ \alpha = \frac{1}{2}, \quad \beta = 1, \quad \gamma = -\frac{1}{2} \] Using the distance formula: \[ d = \frac{|3(3) + 2(1) + (-1)(-1) + 29|}{\sqrt{3^2 + 2^2 + (-1)^2}} \] \[ = \frac{42}{\sqrt{14}} = 3\sqrt{14} \]