A line 'ℓ' passing through origin is perpendicular to the lines ℓ1 : r = (3 + t)i + (-1 + 2t)j + (4

Q: Why did we use the cross product to find the direction of $\ell$?

The cross product $\vec{d_1} \times \vec{d_2}$ is the fundamental tool in 3D geometry to find a vector perpendicular to two given vectors. Since $\ell$ is perpendicular to both $\ell_1$ and $\ell_2$, its direction must align with this cross product. It is the geometric equivalent of finding the 'normal' to the plane containing both lines.

Q: Why do we reject $s = -2$?

The problem explicitly specifies that point $Q$ lies in the first octant. This means all coordinates $(a, b, c)$ must be strictly positive. When we plug $s = -2$ into the coordinates of $Q$, we get $(-1, -1, 0)$, which fails this condition. Always check your constraints at the end of the problem; they are the gatekeepers of your final answer.

Q: Is there a faster way to find the intersection point $P$?

While you could use the shortest distance formula between skew lines, finding the intersection point by equating the parametric forms is often more robust and less prone to calculation errors in this specific setup. It gives you a clear, step-by-step path to the coordinates.

Question

A line '$\ell$' passing through origin is perpendicular to the lines $\ell_1 : \vec{r} = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$ and $\ell_2 : \vec{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$. If the co-ordinates of the point in the first octant on '$\ell_2$' at the distance of $\sqrt{17}$ from the point of intersection of '$\ell$' and '$\ell_1$' are $(a, b, c)$, then $18(a + b + c)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE Advanced landscape. Today, we are not just solving a problem; we are navigating the intricate architecture of 3D space.

We have two lines, $\ell_1$ and $\ell_2$, drifting through the coordinate system, and a third line, $\ell$, that acts as a bridge between them. This is a classic problem that tests your ability to visualize vectors and enforce geometric constraints.

The equations for the lines are given in vector form:
$$\ell_1 : \vec{r} = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$$
$$\ell_2 : \vec{r} = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$$

The coefficients of the parameters $t$ and $s$ define the direction vectors $\vec{d_1} = (1, 2, 2)$ and $\vec{d_2} = (2, 2, 1)$.

### The Direction of the Unknown

We introduce line $\ell$, which is perpendicular to both $\ell_1$ and $\ell_2$. In the language of vectors, if a vector is perpendicular to two others, it must be parallel to their cross product.

We calculate the direction vector $\vec{d} = \vec{d_1} 	imes \vec{d_2}$:
$$\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2 - 4) - \hat{j}(1 - 4) + \hat{k}(2 - 4) = -2\hat{i} + 3\hat{j} - 2\hat{k}$$

This vector $\vec{d} = (-2, 3, -2)$ is the compass that guides our line $\ell$ through the origin $(0, 0, 0)$.

### The Intersection Point $P$

With the direction vector $(-2, 3, -2)$ and the origin $(0, 0, 0)$, the equation of line $\ell$ is:
$$\frac{x}{-2} = \frac{y}{3} = \frac{z}{-2} = k$$

Any point on this line can be represented as $(-2k, 3k, -2k)$. We find the intersection point $P$ between $\ell$ and $\ell_1$ by equating their coordinates:
$$-2k = 3 + t$$
$$-2k = 4 + 2t$$

Equating these expressions, we get $3 + t = 4 + 2t$, which yields $t = -1$. Substituting $t = -1$ back into the equation for $\ell_1$, we find the coordinates of $P$ to be $(2, -3, 2)$.

### The Hunt for $Q$ and the Octant Constraint

Point $Q$ lies on $\ell_2$ with coordinates $(3 + 2s, 3 + 2s, 2 + s)$. We are given that the distance $PQ = \sqrt{17}$, so $PQ^2 = 17$:
$$(3 + 2s - 2)^2 + (3 + 2s - (-3))^2 + (2 + s - 2)^2 = 17$$

Expanding this, we obtain:
$$(2s + 1)^2 + (2s + 6)^2 + s^2 = 17$$
$$4s^2 + 4s + 1 + 4s^2 + 24s + 36 + s^2 = 17$$
$$9s^2 + 28s + 20 = 0$$

Solving the quadratic equation $9s^2 + 28s + 20 = 0$ gives:
$$(9s + 10)(s + 2) = 0 \Rightarrow s = -2, s = -\frac{10}{9}$$

We must satisfy the 'first octant' condition, where $x, y, z > 0$. Testing $s = -2$ yields $Q = (-1, -1, 0)$, which is invalid. Testing $s = -\frac{10}{9}$ yields:
$$Q = \left(3 - \frac{20}{9}, 3 - \frac{20}{9}, 2 - \frac{10}{9}ight) = \left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}ight)$$

### Final Calculation

The problem asks for $18(a + b + c)$, where $(a, b, c)$ are the coordinates of $Q$. The sum is:
$$a + b + c = \frac{7}{9} + \frac{7}{9} + \frac{8}{9} = \frac{22}{9}$$

Multiplying by $18$:
$$18 	imes \frac{22}{9} = 2 	imes 22 = 44$$

The final answer is **44**.

Enter Numerical Value:

Visualized Solution

Identify Direction Vectors of $ℓ_{1}$ and $ℓ_{2}$

Find the Direction of Line $ℓ$

Equation of Line $ℓ$

Finding Intersection Point $P$

Coordinates of Point $P$

Define Point $Q$ on $ℓ_{2}$

Apply Distance Formula $P Q = 17$

Expand and Simplify the Equation

Solve the Quadratic Equation

Check First Octant Condition

Calculate Final Expression

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Direction of the Unknown

The Intersection Point $P$

The Hunt for $Q$ and the Octant Constraint

Final Calculation

Similar Questions

Let $A$ and $B$ be two distinct points on the line $L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Both $A$ and $B$ are at a distance $217$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$ . If $O$ is the origin, then $O A \cdot O B$ is equal to:

Let a line passing through the point $(4, 1, 0)$ intersect the line $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ at the point $A (α, β, γ)$ and the line $L_{2} : x - 6 = y = - z + 4$ at the point $B (a, b, c)$ . Then $1 α a 0 β b 1 γ c$ is equal to

If the distances of the point $(1, 2, a)$ from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and $L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to

Enter Numerical Value:

Visualized Solution

Identify Direction Vectors of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ℓ1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ℓ2​

Find the Direction of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ℓ

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ℓ

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Define Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ℓ2​

Apply Distance Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ=17​