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3D Geometry question

2025 · 4 Apr · Shift 1 · Q27

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JEE Main
Mathematics
3D Geometry
2025 · 4 Apr · Shift 1 · Q27

3D Geometry question

2025 · 4 Apr · Shift 1 · Q27

JEE MainMathematics3D GeometryMCQ+4 / −1

Let the shortest distance between the lines

\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}

and

\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}

be

3 \sqrt{30}

. Then the positive value of

5 \alpha+\beta

is

A
42
B
40
C
48
D
46

View written solutionFree

Correct answer: D

Write the lines in vector form

The given lines are

$\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}.$

So we can identify:

A point on line $L_1$ : $A=(3,\alpha,3)$
Direction vector of $L_1$ : $\vec d_1=(3,-1,1)$
A point on line $L_2$ : $B=(-3,-7,\beta)$
Direction vector of $L_2$ : $\vec d_2=(-3,2,4)$

Use formula for shortest distance between two skew lines

For lines through points $A,B$ with direction vectors $\vec d_1,\vec d_2$ , the shortest distance is

$D=\frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Given $D=3\sqrt{30}$ .

Compute the cross product

\begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -1 & 1\\ -3 & 2 & 4 \end{vmatrix}$$ $$=\hat i((-1)(4)-(1)(2)) - \hat j((3)(4)-(1)(-3)) + \hat k((3)(2)-(-1)(-3))$$ $$=\hat i(-4-2)-\hat j(12+3)+\hat k(6-3)$$ $$=(-6,-15,3).$$ Hence, $$|\vec d_1\times \vec d_2|=\sqrt{(-6)^2+(-15)^2+3^2} =\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}.$$ --- 4. **Use the distance value** Since the denominator is already $3\sqrt{30}$ and the shortest distance is also $3\sqrt{30}$, $$\frac{|\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)|}{3\sqrt{30}}=3\sqrt{30}.$$ Therefore, $$|\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)|=(3\sqrt{30})(3\sqrt{30})=270.$$ --- 5. **Compute $\overrightarrow{AB}$** $$\overrightarrow{AB}=B-A=(-3-3, -7-\alpha, \beta-3)=(-6,-7-\alpha,\beta-3).$$ Now, $$\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2) =(-6,-7-\alpha,\beta-3)\cdot(-6,-15,3).$$ So, $$= (-6)(-6)+(-7-\alpha)(-15)+3(\beta-3)$$ $$=36+105+15\alpha+3\beta-9$$ $$=132+15\alpha+3\beta.$$ Thus, $$|132+15\alpha+3\beta|=270.$$ Divide by $3$: $$|44+5\alpha+\beta|=90.$$ So, $$44+5\alpha+\beta=\pm 90.$$ Hence, $$5\alpha+\beta=46 \quad \text{or} \quad -134.$$ The question asks for the **positive value** of $5\alpha+\beta$. Therefore, $$5\alpha+\beta=46.$$ --- 6. **Check options** - A: $42$ ❌ - B: $40$ ❌ - C: $48$ ❌ - D: $46$ ✅ So the correct option is **D**.

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