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Step 1: Express Each Line in Parametric Form
• First line:
\frac{x-2}{2} = \frac{y}{-2} = \frac{z-7}{16} .
Let \lambda be the parameter. Then:
x = 2 + 2\lambda,\quad
y = -2\lambda,\quad
z = 7 + 16\lambda.
• Second line:
\frac{x+3}{4} = \frac{y+2}{3} = \frac{z+2}{1} .
Let \mu be the parameter. Then:
x = -3 + 4\mu,\quad
y = -2 + 3\mu,\quad
z = -2 + \mu.
Step 2: Find the Intersection Point P
At the intersection point, the coordinates must satisfy both sets of parametric equations simultaneously, giving us the system:
2 + 2\lambda = -3 + 4\mu
-2\lambda = -2 + 3\mu
7 + 16\lambda = -2 + \mu
From the second equation:
-2\lambda = -2 + 3\mu
\;\Longrightarrow\; 3\mu = -2\lambda + 2
\;\Longrightarrow\; \mu = \frac{-2\lambda + 2}{3}.
Substitute \mu = \tfrac{-2\lambda + 2}{3} into the first equation:
2 + 2\lambda = -3 + 4 \left(\frac{-2\lambda + 2}{3}\right).
Simplify:
2 + 2\lambda = -3 + \frac{-8\lambda + 8}{3}.
Bring terms together:
5 + 2\lambda = \frac{-8\lambda + 8}{3}.
Multiply by 3:
15 + 6\lambda = -8\lambda + 8
\;\Longrightarrow\; 14\lambda = -7
\;\Longrightarrow\; \lambda = -\frac{1}{2}.
Now substitute \lambda = -\tfrac{1}{2} back into \mu = \tfrac{-2\lambda + 2}{3} :
\mu = \frac{-2 \left(-\frac{1}{2}\right) + 2}{3} = \frac{1 + 2}{3} = 1.
Step 3: Calculate the Coordinates of P
Substitute \lambda = -\frac{1}{2} into the parametric form of the first line:
x_P = 2 + 2 \left(-\frac{1}{2}\right) = 2 - 1 = 1.
y_P = -2 \left(-\frac{1}{2}\right) = 1.
z_P = 7 + 16 \left(-\frac{1}{2}\right) = 7 - 8 = -1.
Hence, P(1,\,1,\,-1) is the intersection point.
Step 4: Write the Third Line in Parametric Form
The third line:
\frac{x+1}{2} = \frac{y-1}{3} = \frac{z-1}{1} .
Let t be the parameter. Then we get:
x = -1 + 2t,
y = 1 + 3t,
z = 1 + t.
A convenient point on this line is A(-1,\,1,\,1) , and the direction vector is
\vec{d} = (2,\,3,\,1).
Step 5: Calculate the Distance of P from the Third Line
The distance l from a point P to a line passing through a point A with direction
\vec{d} is given by:
l = \frac{\left\lvert \overrightarrow{AP} \times \vec{d} \right\rvert}{\left\lvert \vec{d} \right\rvert}.
Compute \overrightarrow{AP} = \bigl(1 - (-1),\; 1 - 1,\; -1 - 1\bigr) = (2,\,0,\,-2).
Compute the cross product \overrightarrow{AP} \times \vec{d} where \vec{d} = (2,\,3,\,1) :
\overrightarrow{AP} \times \vec{d}
= \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 0 & -2 \\
2 & 3 & 1
\end{vmatrix}
= \mathbf{i} \bigl(0 \cdot 1 - (-2)\cdot 3 \bigr)
- \mathbf{j} \bigl(2 \cdot 1 - (-2)\cdot 2 \bigr)
+ \mathbf{k} \bigl(2 \cdot 3 - 0 \cdot 2 \bigr).
= (6)\mathbf{i}
- (2 + 4)\mathbf{j}
+ (6)\mathbf{k}
= (6,\,-6,\,6).
Find the magnitude \bigl|\overrightarrow{AP} \times \vec{d}\bigr| :
|(6, -6, 6)| = \sqrt{6^2 + (-6)^2 + 6^2} = \sqrt{36 + 36 + 36} = \sqrt{108} = 6\sqrt{3}.
Magnitude of the direction vector \vec{d} = (2,\,3,\,1) :
|(2, 3, 1)| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}.
Therefore, the distance l from P to the line is:
l = \frac{6\sqrt{3}}{\sqrt{14}}.
Step 6: Compute 14\,l^2
First, find l^2 :
l^2 = \left(\frac{6\sqrt{3}}{\sqrt{14}}\right)^2
= \frac{36 \cdot 3}{14}
= \frac{108}{14}.
Then multiply by 14:
14\,l^2 = 14 \times \frac{108}{14} = 108.
Final Answer
14\,l^2 = 108.
