Let a Straight Line L Pass Through P(2, −1, 3) and Be Perpendicular to the Lines (x−1)/2 = (y+1)/1 = (z−3)/(−2) and (x−3)/1 = (y−2)/3 = (z+2)/4. If L Intersects the yz-Plane at Q, Then the Distance Between P and Q Is

JEE Main 2025 — 29 January, Evening Shift. Previous Year Question.

Problem

Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-3}{-2}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{3}=\dfrac{z+2}{4}$. If the line $L$ intersects the yz-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

Key insight. Being perpendicular to two given lines pins down $L$’s direction uniquely — it must be parallel to the cross product of the two lines’ direction vectors, since the cross product is perpendicular to both simultaneously. From there, finding where $L$ crosses the yz-plane is just a matter of setting $x=0$ in its parametric equation.

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Approach

Find the direction vector of $L$ as the cross product of the two given lines’ direction vectors. Write $L$’s parametric equation using this direction and the point $P$. Set the $x$-coordinate to zero to find the parameter value where $L$ meets the yz-plane, giving the point $Q$, then compute the distance $PQ$.

Solution

Step 1 — Identify the direction vectors of the two given lines

$$\vec{d_1} = (2,1,-2), \qquad \vec{d_2} = (1,3,4)$$

Step 2 — Compute the direction of L as d₁ × d₂

$$\vec{d_1}\times\vec{d_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-2\\1&3&4\end{vmatrix}$$

$$= \hat{i}\big[(1)(4)-(-2)(3)\big] – \hat{j}\big[(2)(4)-(-2)(1)\big] + \hat{k}\big[(2)(3)-(1)(1)\big]$$

$$= \hat{i}(4+6) – \hat{j}(8+2) + \hat{k}(6-1) = (10,-10,5)$$

Simplifying by dividing by 5: direction $(2,-2,1)$.

Step 3 — Write the parametric equation of L

$$(x,y,z) = (2,-1,3) + t(2,-2,1) = (2+2t,\ -1-2t,\ 3+t)$$

Step 4 — Find where L meets the yz-plane (x = 0)

$$2+2t = 0 \implies t=-1$$

Step 5 — Find Q

$$Q = (2+2(-1),\ -1-2(-1),\ 3+(-1)) = (0,\ 1,\ 2)$$

Step 6 — Compute the distance PQ

$$PQ = \sqrt{(2-0)^2+(-1-1)^2+(3-2)^2} = \sqrt{4+4+1} = \sqrt9 = 3$$

Answer

Common mistakes

• Trying to find $L$’s direction using only one of the two given lines. A single perpendicularity condition leaves infinitely many possible directions (an entire plane of them) — both conditions together are needed, combined via the cross product, to pin down a single direction.
• Setting the wrong coordinate to zero for the yz-plane. The yz-plane is where $x=0$ (not $y=0$ or $z=0$) — confusing this with the xz-plane or xy-plane leads to solving for the wrong parameter value entirely.

Practise next

• Let a line $M$ pass through $(1,2,-1)$, perpendicular to $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z+2}{-1}$ and $\dfrac{x-2}{2}=\dfrac{y}{-1}=\dfrac{z+1}{1}$. Find where $M$ meets the xy-plane and the distance from the original point, using the same cross-product method.