If the Length of the Perpendicular From the Point (β, 0, β), β ≠ 0, to the Line x/1 = (y−1)/0 = (z+1)/(−1) Is √(3/2), Then β Is Equal To

JEE Main 2019 — 10 April, Morning Slot. Previous Year Question.

Problem

If the length of the perpendicular from the point $(\beta,0,\beta)$, $\beta\neq0$, to the line $\dfrac{x}{1}=\dfrac{y-1}{0}=\dfrac{z+1}{-1}$ is $\sqrt{\dfrac{3}{2}}$, then $\beta$ is equal to:

Key insight. The perpendicular distance from a point to a line in 3D is found using the cross product of the direction vector with the vector from a known point on the line to the given point — a direct formula that avoids constructing the actual foot of the perpendicular.

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Approach

Identify a point $A$ on the line and its direction vector $\vec{d}$. Form the vector $\vec{AP}$ from $A$ to the given point $P=(\beta,0,\beta)$, compute $\vec{AP}\times\vec{d}$, and use the formula $\text{distance} = \dfrac{|\vec{AP}\times\vec{d}|}{|\vec{d}|}$. Set this equal to the given distance and solve for $\beta$.

Solution

Step 1 — Identify a point on the line and its direction vector

The line $\dfrac{x}{1}=\dfrac{y-1}{0}=\dfrac{z+1}{-1}$ passes through $A=(0,1,-1)$ with direction $\vec{d}=(1,0,-1)$.

Step 2 — Form the vector from A to P

With $P=(\beta,0,\beta)$:

$$\vec{AP} = (\beta-0,\ 0-1,\ \beta-(-1)) = (\beta,\ -1,\ \beta+1)$$

Step 3 — Compute AP × d

$$\vec{AP}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\beta&-1&\beta+1\\1&0&-1\end{vmatrix}$$

$$= \hat{i}\big[(-1)(-1)-(\beta+1)(0)\big] – \hat{j}\big[(\beta)(-1)-(\beta+1)(1)\big] + \hat{k}\big[(\beta)(0)-(-1)(1)\big]$$

$$= \hat{i}(1) – \hat{j}(-\beta-\beta-1) + \hat{k}(1) = \hat{i} + (2\beta+1)\hat{j} + \hat{k}$$

Step 4 — Compute the magnitudes

$$|\vec{AP}\times\vec{d}| = \sqrt{1+(2\beta+1)^2+1} = \sqrt{2+(2\beta+1)^2}$$

$$|\vec{d}| = \sqrt{1+0+1} = \sqrt2$$

Step 5 — Apply the distance formula and set it equal to the given value

$$\frac{\sqrt{2+(2\beta+1)^2}}{\sqrt2} = \sqrt{\frac{3}{2}}$$

Step 6 — Square both sides

$$\frac{2+(2\beta+1)^2}{2} = \frac{3}{2} \implies 2+(2\beta+1)^2 = 3 \implies (2\beta+1)^2=1$$

Step 7 — Solve for β

$$2\beta+1 = \pm1$$

Taking $+1$: $2\beta=0 \implies \beta=0$ — rejected, since $\beta\neq0$ is given.

Taking $-1$: $2\beta=-2 \implies \beta=-1$

Answer

Common mistakes

• Forgetting to reject $\beta=0$. The algebra naturally produces two solutions, but the problem explicitly excludes $\beta=0$ — missing this leaves an invalid extra “answer” that isn’t among the intended options anyway.
• Mixing up which point is $A$ (on the line) and which is $P$ (off the line) when forming the vector for the cross product. Reversing them flips the sign of the resulting vector, but since the formula uses magnitudes, this particular slip is harmless here — though it’s still worth being consistent to avoid confusion in more involved problems.

Practise next

• Find the length of the perpendicular from the point $(2,1,-1)$ to the line $\dfrac{x-1}{2}=\dfrac{y}{1}=\dfrac{z+1}{-2}$, using the same cross-product distance formula.