If the shortest distance between the lines \overrightarrow r = ( - \widehat i + 3\widehat k ) + \lambda ( \widehat i - a\widehat j ) and \overrightarrow r = ( - \widehat j + 2\widehat k ) + \mu ( \widehat i - \widehat j + \widehat k ) is \sqrt 2 \over 3 , then the integral value of a is equal to \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2022 · 24 Jun · Shift 1 · Q43 with Solution

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Correct answer: 2

Write the lines in standard vector form

The given lines are $L_1: \vec r = (-\hat i+3\hat k)+\lambda(\hat i-a\hat j)$ and $L_2: \vec r = (-\hat j+2\hat k)+\mu(\hat i-\hat j+\hat k).$

So, points on the lines are: $A=(-1,0,3), \qquad B=(0,-1,2).$

Their direction vectors are: $\vec d_1=(1,-a,0), \qquad \vec d_2=(1,-1,1).$

Use the formula for shortest distance between two skew lines

For lines passing through points $A,B$ with direction vectors $\vec d_1,\vec d_2$ , the shortest distance is $D=\frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here, $\overrightarrow{AB}=B-A=(0-(-1),-1-0,2-3)=(1,-1,-1).$

Given shortest distance: $D=\sqrt{\frac23}.$

Compute the cross product $\vec d_1\times \vec d_2$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -a & 0\\ 1 & -1 & 1 \end{vmatrix}$$ $$=\hat i((-a)(1)-0(-1)) - \hat j(1\cdot 1-0\cdot 1)+\hat k(1(-1)-(-a)(1))$$ $$=(-a)\hat i-\hat j+(a-1)\hat k.$$ So, $$\vec d_1\times \vec d_2=(-a,-1,a-1).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{a^2+1+(a-1)^2}.$$ Now, $$(a-1)^2=a^2-2a+1,$$ so $$|\vec d_1\times \vec d_2|=\sqrt{a^2+1+a^2-2a+1}= \sqrt{2a^2-2a+2}.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2) =(1,-1,-1)\cdot(-a,-1,a-1).$$ $$=1(-a)+(-1)(-1)+(-1)(a-1)$$ $$=-a+1-a+1=2-2a=2(1-a).$$ Hence, $$|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|=|2(1-a)|=2|1-a|.$$ --- 5. **Set up the distance equation** $$\frac{2|1-a|}{\sqrt{2a^2-2a+2}}=\sqrt{\frac23}.$$ Square both sides: $$\frac{4(1-a)^2}{2a^2-2a+2}=\frac23.$$ Cross-multiply: $$12(1-a)^2=2(2a^2-2a+2).$$ $$6(1-a)^2=2a^2-2a+2.$$ Now expand: $$6(a^2-2a+1)=2a^2-2a+2.$$ $$6a^2-12a+6=2a^2-2a+2.$$ $$4a^2-10a+4=0.$$ $$2a^2-5a+2=0.$$ Factorizing: $$2a^2-5a+2=(2a-1)(a-2)=0.$$ Thus, $$a=\frac12 \quad \text{or} \quad a=2.$$ --- 6. **Choose the integral value** The question asks for the **integral value** of $a$. Hence, $$\boxed{a=2}.$$ --- 7. **Comparison with stored answer** Stored correct answer: $2$ Our derived answer is also $2$, so they agree.

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View written solutionFree

Correct answer: 2

Write the lines in standard vector form

The given lines are $L_1: \vec r = (-\hat i+3\hat k)+\lambda(\hat i-a\hat j)$ and $L_2: \vec r = (-\hat j+2\hat k)+\mu(\hat i-\hat j+\hat k).$

So, points on the lines are: $A=(-1,0,3), \qquad B=(0,-1,2).$

Their direction vectors are: $\vec d_1=(1,-a,0), \qquad \vec d_2=(1,-1,1).$

Use the formula for shortest distance between two skew lines

Here, $\overrightarrow{AB}=B-A=(0-(-1),-1-0,2-3)=(1,-1,-1).$

Given shortest distance: $D=\sqrt{\frac23}.$

Compute the cross product $\vec d_1\times \vec d_2$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -a & 0\\ 1 & -1 & 1 \end{vmatrix}$$ $$=\hat i((-a)(1)-0(-1)) - \hat j(1\cdot 1-0\cdot 1)+\hat k(1(-1)-(-a)(1))$$ $$=(-a)\hat i-\hat j+(a-1)\hat k.$$ So, $$\vec d_1\times \vec d_2=(-a,-1,a-1).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{a^2+1+(a-1)^2}.$$ Now, $$(a-1)^2=a^2-2a+1,$$ so $$|\vec d_1\times \vec d_2|=\sqrt{a^2+1+a^2-2a+1}= \sqrt{2a^2-2a+2}.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2) =(1,-1,-1)\cdot(-a,-1,a-1).$$ $$=1(-a)+(-1)(-1)+(-1)(a-1)$$ $$=-a+1-a+1=2-2a=2(1-a).$$ Hence, $$|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|=|2(1-a)|=2|1-a|.$$ --- 5. **Set up the distance equation** $$\frac{2|1-a|}{\sqrt{2a^2-2a+2}}=\sqrt{\frac23}.$$ Square both sides: $$\frac{4(1-a)^2}{2a^2-2a+2}=\frac23.$$ Cross-multiply: $$12(1-a)^2=2(2a^2-2a+2).$$ $$6(1-a)^2=2a^2-2a+2.$$ Now expand: $$6(a^2-2a+1)=2a^2-2a+2.$$ $$6a^2-12a+6=2a^2-2a+2.$$ $$4a^2-10a+4=0.$$ $$2a^2-5a+2=0.$$ Factorizing: $$2a^2-5a+2=(2a-1)(a-2)=0.$$ Thus, $$a=\frac12 \quad \text{or} \quad a=2.$$ --- 6. **Choose the integral value** The question asks for the **integral value** of $a$. Hence, $$\boxed{a=2}.$$ --- 7. **Comparison with stored answer** Stored correct answer: $2$ Our derived answer is also $2$, so they agree.

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