The shortest distance between the lines x-1/2 = y-2/3 = z+4/6 and x-3/2 = y-3/3 = z+5/6 is . . . . . . .

Correct Answer is : 29349{293/49}49293

Correct Answer is : 29349{293/49}49293

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GUJCET 2025 Math Paper

Question : 3 of 40

Marks: +1, -0

\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}

\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}

is . . . . . . .

Solution:

Concept:

For parallel lines

\vec{r}=\vec{a}_1+\lambda \vec{b}

and

\vec{r}=\vec{a}_2+\mu \vec{b}

, the shortest distance is

d=\dfrac{|(\vec{a}_2-\vec{a}_1)\times \vec{b}|}{|\vec{b}|}

Explanation:

Direction vector is the same:

\vec{b}=(2,3,6)

Points on the lines are

\vec{a}_1=(1,2,-4)

and

\vec{a}_2=(3,3,-5)

So,

\vec{a}_2-\vec{a}_1=(2,1,-1)

Now,

(\vec{a}_2-\vec{a}_1)\times \vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\2&3&6\end{vmatrix}=9\hat{i}-14\hat{j}+4\hat{k}

Its magnitude is

|(\vec{a}_2-\vec{a}_1)\times \vec{b}|=\sqrt{9^2+(-14)^2+4^2}=\sqrt{293}

Also,

|\vec{b}|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7

Thus,

d=\dfrac{\sqrt{293}}{7}=\sqrt{\dfrac{293}{49}}

Answer:

\sqrt{\dfrac{293}{49}}

(Option B).

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