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Step-by-Step Solution
Step 1: Determine the Direction Vectors of lā and lā
The given lines are:
ā¢ l_{1}: \vec{r} = \langle 1,\,-11,\,-7\rangle + \lambda \langle 1,\;2,\;3\rangle
ā¢ l_{2}: \vec{r} = \langle -1,\;0,\;1\rangle + \mu \langle 2,\;2,\;1\rangle
Therefore, the direction vector of l_{1} is \langle 1,\;2,\;3\rangle , and the direction vector of l_{2} is \langle 2,\;2,\;1\rangle .
Step 2: Find the Direction Vector of Line l
Line l is perpendicular to both l_{1} and l_{2} . Hence, its direction vector is the cross product of the direction vectors of l_{1} and l_{2} :
\langle a,\,b,\,c\rangle
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\
2 & 2 & 1
\end{vmatrix}.
Step 3: Compute the Cross Product
Evaluate the determinant:
\langle a,\,b,\,c\rangle
=
\hat{i}\,(2 \cdot 1 - 3 \cdot 2)
-
\hat{j}\,(1 \cdot 1 - 3 \cdot 2)
+
\hat{k}\,(1 \cdot 2 - 2 \cdot 2).
i-component: 2 \cdot 1 - 3 \cdot 2 = 2 - 6 = -4.
j-component: 1 \cdot 1 - 3 \cdot 2 = 1 - 6 = -5, but there is a minus sign in front (from the determinant expansion), so it becomes +5.
k-component: 1 \cdot 2 - 2 \cdot 2 = 2 - 4 = -2.
Thus, the direction vector of l is \langle -4,\;5,\;-2\rangle .
Step 4: Write the Equation of Line l
Since l passes through the origin (0,0,0) and has direction \langle -4,\;5,\;-2\rangle , its vector equation is:
\vec{r} = \delta \,\langle -4,\;5,\;-2\rangle.
Equivalently, in parametric form:
x = -4\,\delta,\quad y = 5\,\delta,\quad z = -2\,\delta.
Step 5: Find Intersection Point P of l and lā
The line l_{1} is:
\vec{r}
=
\langle 1,\;-11,\;-7\rangle
+
\lambda \,\langle 1,\,2,\,3\rangle.
At the intersection point P , the coordinates on l and l_{1} are equal. Hence:
\langle -4\delta,\;5\delta,\;-2\delta\rangle
=
\langle 1 + \lambda,\;-11 + 2\lambda,\;-7 + 3\lambda\rangle.
This gives the system of equations:
-4\delta = 1 + \lambda,\quad
5\delta = -11 + 2\lambda,\quad
-2\delta = -7 + 3\lambda.
Step 6: Solve for Ī“ and Ī»
From -4\delta = 1 + \lambda, express \lambda in terms of \delta and substitute into the other equations. Upon simplifying, you find:
\delta = -1,\quad \lambda = 3.
Substituting \delta = -1 into \vec{r} = \delta\,\langle -4,\;5,\;-2\rangle yields:
P = \langle -4(-1),\;5(-1),\;-2(-1)\rangle = \langle 4,\;-5,\;2\rangle.
Step 7: Parametric Form of Points on lā
The line l_{2} is:
\vec{r}
=
\langle -1,\;0,\;1\rangle
+
\mu\,\langle 2,\;2,\;1\rangle.
Any point Q on l_{2} can be written as:
Q(\mu) = \langle -1 + 2\mu,\;0 + 2\mu,\;1 + \mu\rangle.
Step 8: Foot of the Perpendicular Condition
The point Q(\mu) is the foot of the perpendicular from P onto l_{2} , so the vector \overrightarrow{PQ} is perpendicular to the direction \langle 2,\;2,\;1\rangle of l_{2} . Hence:
\overrightarrow{PQ} \cdot \langle 2,\;2,\;1\rangle = 0.
First, find \overrightarrow{PQ} :
\overrightarrow{PQ}
=
\langle (-1 + 2\mu) - 4,\;(2\mu) - (-5),\;(1 + \mu) - 2\rangle
=
\langle -5 + 2\mu,\;2\mu + 5,\;\mu - 1\rangle.
Now take the dot product with \langle 2,\,2,\,1\rangle and set it to 0:
2(-5 + 2\mu) + 2(2\mu + 5) + 1(\mu - 1) = 0.
Simplify:
-10 + 4\mu + 4\mu + 10 + \mu - 1 = 0
\quad\Longrightarrow\quad
9\mu - 1 = 0
\quad\Longrightarrow\quad
\mu = \frac{1}{9}.
Step 9: Coordinates of Q
Substitute \mu = \tfrac{1}{9} back into Q(\mu) :
\alpha = -1 + \frac{2}{9} = -\frac{7}{9},\quad
\beta = 2 \times \frac{1}{9} = \frac{2}{9},\quad
\gamma = 1 + \frac{1}{9} = \frac{10}{9}.
Step 10: Calculate 9(\alpha + \beta + \gamma)
Sum these coordinates:
\alpha + \beta + \gamma
=
\left(-\frac{7}{9}\right) + \frac{2}{9} + \frac{10}{9} = \frac{5}{9}.
Therefore,
9(\alpha + \beta + \gamma) = 9 \times \frac{5}{9} = 5.
Final Answer
The value of 9(\alpha + \beta + \gamma) is \boxed{5} .
