© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the direction ratios (D.R.) of the given lines
Line l₁ is given by
\overrightarrow{r} = (3 + t)\,\hat{i} + (-1 + 2t)\,\hat{j} + (4 + 2t)\,\hat{k}.
From its symmetric form
\frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 4}{2},
the direction ratios (D.R.) of l₁ are
(1,\,2,\,2).
Line l₂ is given by
\overrightarrow{r} = (3 + 2s)\,\hat{i} + (3 + 2s)\,\hat{j} + (2 + s)\,\hat{k}.
From its symmetric form
\frac{x - 3}{2} = \frac{y - 3}{2} = \frac{z - 2}{1},
the direction ratios (D.R.) of l₂ are
(2,\,2,\,1).
Step 2: Determine the direction ratios of line l
The line l passes through the origin and is perpendicular to both l₁ and l₂. A line that is perpendicular to two lines in 3D has a direction vector parallel to the cross product of their direction vectors. Hence, if
\text{D.R. of } l₁ = (1,\,2,\,2)\quad\text{and}\quad \text{D.R. of } l₂ = (2,\,2,\,1),
then
\text{D.R. of } l \propto (1,\,2,\,2) \times (2,\,2,\,1).
Compute the cross product:
(1,\,2,\,2) \times (2,\,2,\,1)
= \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 2 \\
2 & 2 & 1
\end{vmatrix}
= \hat{i}(2 \cdot 1 - 2 \cdot 2) - \hat{j}(1 \cdot 1 - 2 \cdot 2) + \hat{k}(1 \cdot 2 - 2 \cdot 2).
Thus:
(1,\,2,\,2) \times (2,\,2,\,1) = (-2,\,3,\,-2).
We can choose a positive scalar multiple to simplify signs. Taking (2,\, -3,\,2) gives the same direction for l.
Step 3: Parametric equation of line l
Since l passes through the origin and has direction ratios (2,\,-3,\,2) , its parametric form can be written as:
x = 2\lambda,\quad y = -3\lambda,\quad z = 2\lambda.
Equivalent symmetric form is
\frac{x}{2} = \frac{y}{-3} = \frac{z}{2}.
Step 4: Find the intersection point of lines l and l₁
Let the point of intersection be
(2\lambda,\,-3\lambda,\,2\lambda)
on l and also
(3 + t,\, -1 + 2t, \,4 + 2t)
on l₁. We must have:
2\lambda = 3 + t, \quad -3\lambda = -1 + 2t, \quad 2\lambda = 4 + 2t.
From 2\lambda = 3 + t and 2\lambda = 4 + 2t, we solve for t and \lambda :
• From 2\lambda = 3 + t → (1)
• From 2\lambda = 4 + 2t → (2)
Compare (1) and (2):
(1) implies t = 2\lambda - 3.
(2) implies 2\lambda = 4 + 2t \implies \lambda = 2 + t.
Setting t = 2\lambda - 3 into \lambda = 2 + t :
\lambda = 2 + (2\lambda - 3) \implies \lambda = 2\lambda - 1 \implies \lambda = 1, \quad t = 2(1) - 3 = -1.
Hence, the point of intersection is
P(2\cdot1,\,-3\cdot1,\,2\cdot1) = (2,\,-3,\,2).
Step 5: General point on l₂ and distance condition
A general point on l₂ is
Q(3 + 2s,\,3 + 2s,\,2 + s),
which can also be written as
(2s + 3,\,2s + 3,\,s + 2).
We want the distance between P(2,\,-3,\,2) and Q(2s+3,\,2s+3,\,s+2) to be \sqrt{17}. Thus:
PQ = \sqrt{(2s + 3 - 2)^2 + (2s + 3 + 3)^2 + (s + 2 - 2)^2} = \sqrt{17}.
So,
(2s + 1)^2 + (2s + 6)^2 + (s)^2 = 17.
Step 6: Solve the quadratic equation for s
Expand:
(2s + 1)^2 = 4s^2 + 4s + 1,
(2s + 6)^2 = 4s^2 + 24s + 36,
(s)^2 = s^2.
Summing these gives:
4s^2 + 4s + 1 \;+\; 4s^2 + 24s + 36 \;+\; s^2 = 17.
Combine like terms:
9s^2 + 28s + 37 = 17 \implies 9s^2 + 28s + 20 = 0.
Factorize if possible:
9s^2 + 18s + 10s + 20 = 0 \implies (9s + 10)(s + 2) = 0.
Thus, s = -\frac{10}{9} \text{ or } s = -2.
Step 7: Choose the solution that lies in the first octant
The first octant requires x,\,y,\,z > 0.
• If s = -2, then (2(-2)+3,\,2(-2)+3,\,(-2)+2) = (-1,\,-1,\,0), which is not in the first octant.
• If s = -\frac{10}{9}, then
x = 2\left(-\frac{10}{9}\right)+3 = 3 - \frac{20}{9} = \frac{27 - 20}{9} = \frac{7}{9},
y = 2\left(-\frac{10}{9}\right)+3 = \frac{7}{9}, \quad
z = \left(-\frac{10}{9}\right) + 2 = \frac{8}{9}.
All coordinates \left(\frac{7}{9},\,\frac{7}{9},\,\frac{8}{9}\right) are positive, so this point is indeed in the first octant.
Step 8: Find the required value of 18(a + b + c)
We have
a = \frac{7}{9}, \quad b = \frac{7}{9}, \quad c = \frac{8}{9}.
Hence,
a + b + c = \frac{7}{9} + \frac{7}{9} + \frac{8}{9} = \frac{22}{9}.
Therefore,
18(a + b + c) = 18 \times \frac{22}{9} = 2 \times 22 = 44.
Hence, the required answer is
\boxed{44}.
