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NVAMediumJEE 2023Skew Lines & Shortest Distance

JEE Mathematics 2023 Question with Solution

If the shortest distance between the line joining the points (1,2,3)(1, 2, 3) and (2,3,4)(2, 3, 4), and the line x−12=y+1−1=z−20\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0} is α\alpha, then 28α228\alpha^2 is equal to:

Answer

Correct answer:18

Step-by-step solution

Standard Method

Given: One line joins the points (1,2,3)(1, 2, 3) and (2,3,4)(2, 3, 4). The other line is x−12=y+1−1=z−20\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}. Find: the value of 28α228\alpha^2, where α\alpha is the shortest distance between the two lines.

The shortest distance between two skew lines is

d=∣(b−a)⋅(p×q)∣∣p×q∣d = \frac{|(\mathbf{b}-\mathbf{a}) \cdot (\mathbf{p} \times \mathbf{q})|}{|\mathbf{p} \times \mathbf{q}|}

From the solution,

r⃗=(i^+2j^+3k^)+λ(i^+j^+k^)=a⃗+λp⃗\vec{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(\hat{i}+\hat{j}+\hat{k}) = \vec{a} + \lambda \vec{p} r⃗=(i^−j^+2k^)+μ(2i^−j^)=b⃗+μq⃗\vec{r} = (\hat{i}-\hat{j}+2\hat{k}) + \mu(2\hat{i}-\hat{j}) = \vec{b} + \mu \vec{q}

So,

p⃗=i^+j^+k^,q⃗=2i^−j^,a⃗=i^+2j^+3k^,b⃗=i^−j^+2k^\vec{p} = \hat{i}+\hat{j}+\hat{k}, \quad \vec{q} = 2\hat{i}-\hat{j}, \quad \vec{a} = \hat{i}+2\hat{j}+3\hat{k}, \quad \vec{b} = \hat{i}-\hat{j}+2\hat{k}

Now,

p⃗×q⃗=∣i^j^k^1112−10∣=i^+2j^−3k^\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} = \hat{i}+2\hat{j}-3\hat{k}

Therefore,

d=∣(b⃗−a⃗)⋅(p⃗×q⃗)∣p⃗×q⃗∣∣d = \left|\frac{(\vec{b}-\vec{a}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|}\right| =∣(−3j^−k^)⋅(i^+2j^−3k^)14∣= \left|\frac{(-3\hat{j}-\hat{k}) \cdot (\hat{i}+2\hat{j}-3\hat{k})}{\sqrt{14}}\right| =∣−6+314∣=314= \left|\frac{-6+3}{\sqrt{14}}\right| = \frac{3}{\sqrt{14}}

Hence,

α=314\alpha = \frac{3}{\sqrt{14}}

Now,

28α2=28×914=1828\alpha^2 = 28 \times \frac{9}{14} = 18

Therefore, the required value is 1818.

Using direction vectors of the two lines

Given: The first line passes through (1,2,3)(1,2,3) and (2,3,4)(2,3,4). The second line is x−12=y+1−1=z−20\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}. Find: 28α228\alpha^2.

For the line joining (1,2,3)(1,2,3) and (2,3,4)(2,3,4), the direction vector is obtained from the difference of the two points:

(2−1,3−2,4−3)=(1,1,1)(2-1, 3-2, 4-3) = (1,1,1)

For the second line, the direction ratios are

(2,−1,0)(2,-1,0)

Using the extracted working,

p⃗=(1,1,1),q⃗=(2,−1,0)\vec{p} = (1,1,1), \quad \vec{q} = (2,-1,0)

Their cross product is

p⃗×q⃗=(1,2,−3)\vec{p} \times \vec{q} = (1,2,-3)

with magnitude

∣p⃗×q⃗∣=12+22+(−3)2=14|\vec{p} \times \vec{q}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{14}

A point on the first line is a⃗=(1,2,3)\vec{a} = (1,2,3) and a point on the second line is b⃗=(1,−1,2)\vec{b} = (1,-1,2). Hence,

b⃗−a⃗=(0,−3,−1)\vec{b} - \vec{a} = (0,-3,-1)

Now take the scalar triple product:

(b⃗−a⃗)⋅(p⃗×q⃗)=(0,−3,−1)⋅(1,2,−3)(\vec{b}-\vec{a}) \cdot (\vec{p} \times \vec{q}) = (0,-3,-1) \cdot (1,2,-3) =0−6+3=−3= 0 - 6 + 3 = -3

So the shortest distance is

α=d=∣−3∣14=314\alpha = d = \frac{|{-3}|}{\sqrt{14}} = \frac{3}{\sqrt{14}}

Therefore,

28α2=28×914=1828\alpha^2 = 28 \times \frac{9}{14} = 18

Therefore, the answer is 1818.

Common mistakes

  • Using the wrong direction vector for the line through the two given points. The direction vector must be the difference of the two position vectors, namely (2−1,3−2,4−3)=(1,1,1)(2-1, 3-2, 4-3) = (1,1,1). Do not confuse it with a point on the line.

  • Taking b⃗−a⃗\vec{b}-\vec{a} incorrectly. The shortest-distance formula uses a vector joining any point on one line to any point on the other. Here b⃗−a⃗=(1,−1,2)−(1,2,3)=(0,−3,−1)\vec{b}-\vec{a} = (1,-1,2) - (1,2,3) = (0,-3,-1).

  • Computing p⃗×q⃗\vec{p} \times \vec{q} incorrectly or forgetting its magnitude in the denominator. The numerator uses the scalar triple product, but the denominator must be ∣p⃗×q⃗∣|\vec{p} \times \vec{q}|.

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