The shortest distance between the lines \fracx-44=\fracy+25=\fracz+33 and \fracx-13=\fracy-34=\fracz-42 is :

3D Geometry PYQ — JEE Main 2023 · 8 Apr · Shift 1 · Q28 with Solution

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Correct answer: A

For the line $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}=\lambda,$ we get a point and direction vector: $L_1: \vec r=(4,-2,-3)+\lambda(4,5,3).$ So,

For the line $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}=\mu,$ we get $L_2: \vec r=(1,3,4)+\mu(3,4,2).$ So,

The shortest distance is $d=\frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here, $\overrightarrow{AB}=B-A=(1-4,\,3-(-2),\,4-(-3))=(-3,5,7).$

\begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix}.$$ Expanding, $$\vec d_1\times \vec d_2= \hat i(5\cdot 2-3\cdot 4)-\hat j(4\cdot 2-3\cdot 3)+\hat k(4\cdot 4-5\cdot 3).$$ $$=\hat i(10-12)-\hat j(8-9)+\hat k(16-15)$$ $$=(-2,1,1).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{(-2)^2+1^2+1^2}= \sqrt{4+1+1}=\sqrt{6}.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)=(-3,5,7)\cdot(-2,1,1).$$ $$=(-3)(-2)+5(1)+7(1)=6+5+7=18.$$ --- 5. **Calculate the distance** $$d=\frac{|18|}{\sqrt{6}}=\frac{18}{\sqrt{6}}=\frac{18\sqrt{6}}{6}=3\sqrt{6}.$$ --- 6. **Check options** The value is $$3\sqrt{6},$$ which matches **Option A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** So, they agree.

View written solutionFree

Correct answer: A

For the line $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}=\lambda,$ we get a point and direction vector: $L_1: \vec r=(4,-2,-3)+\lambda(4,5,3).$ So,

For the line $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}=\mu,$ we get $L_2: \vec r=(1,3,4)+\mu(3,4,2).$ So,

The shortest distance is $d=\frac{|(\overrightarrow{AB})\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here, $\overrightarrow{AB}=B-A=(1-4,\,3-(-2),\,4-(-3))=(-3,5,7).$

\begin{vmatrix} \hat i & \hat j & \hat k \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix}.$$ Expanding, $$\vec d_1\times \vec d_2= \hat i(5\cdot 2-3\cdot 4)-\hat j(4\cdot 2-3\cdot 3)+\hat k(4\cdot 4-5\cdot 3).$$ $$=\hat i(10-12)-\hat j(8-9)+\hat k(16-15)$$ $$=(-2,1,1).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{(-2)^2+1^2+1^2}= \sqrt{4+1+1}=\sqrt{6}.$$ --- 4. **Compute the scalar triple product** $$\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)=(-3,5,7)\cdot(-2,1,1).$$ $$=(-3)(-2)+5(1)+7(1)=6+5+7=18.$$ --- 5. **Calculate the distance** $$d=\frac{|18|}{\sqrt{6}}=\frac{18}{\sqrt{6}}=\frac{18\sqrt{6}}{6}=3\sqrt{6}.$$ --- 6. **Check options** The value is $$3\sqrt{6},$$ which matches **Option A**. --- 7. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** So, they agree.