Consider the lines L₁ : x − 1 = y − 2 = z and L₂ : x − 2 = y = z − 1. Let the feet of the perpendiculars from the point P(5, 1, −3) on the lines L₁ and L₂ be Q and R respectively. If the area of the triangle PQR is A, then 4A² is equal to:

 Question:

Consider the lines
$L_{1} : x - 1 = y - 2 = z$
and
$L_{2} : x - 2 = y = z - 1$

Let the feet of the perpendiculars from the point $P(5,\,1,\,-3)$ on the lines $L_{1}$ and $L_{2}$ be $Q$ and $R$ respectively.
If the area of the triangle $PQR$ is $A$, then find $4A^{2}$.

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Question Image

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Short Solution

To find $4A^{2}$, follow these steps:

1. Write lines in parametric form

   • $L_1:x=1+t,y=2+t,z=t$

   • $L_2:x=2+s,y=s,z=1+s$

2. Find the foot of the perpendicular from $P$ to each line
   Use the formula for foot of perpendicular on a line:
   $Q = A + \dfrac{(P-A)\cdot d}{|d|^{2}} d$
   where $A$ is a point on the line and $d$ its direction vector.

3. Get coordinates of $Q$ and $R$ by solving the dot product condition.

4. Find the area of triangle $PQR$
   

   • Compute $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.

   • Area $A = \dfrac12 \|\overrightarrow{PQ}\times\overrightarrow{PR}\|$

5. Finally, calculate $4A^{2}$.

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Image Solution

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Conclusion

After solving:

• $Q$ (foot on $L_1$) comes out as $Q(1,2,0)$
  

• $R$ (foot on $L_2$) is $R(3,1,2)$
  

Vectors:

$$\overrightarrow{PQ} = (-4,1,3),\qquad \overrightarrow{PR} = (-2,0,5)$$

Cross product:

$$\overrightarrow{PQ}\times\overrightarrow{PR} = (5,14,2)$$

Magnitude:

$$\sqrt{5^2+{14}^2+2^2}=\sqrt{25+196+4}=\sqrt{225}=15$$

Area of triangle:

$$A=\frac{1}{2}\times 15=7.5$$

So:

$$4A^2=4\times (7.5{)}^2=4\times 56.25=225$$

✅ Final Answer:

$$\boxed{{\displaystyle 4A^2=\mathrm{225}}}$$

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Video Solution

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