Let the values of p , for which the shortest distance between the lines \fracx+13=\fracy4=\fracz5 and \overrightarrowr=(p \hati+2 \hatj+\hatk)+\lambda(2 \hati+3 \hatj+4 \hatk) is \frac1\sqrt6 , be a, b,(a\lt b) . Then the length of the latus rectum of the ellipse \fracx^2a^2+\fracy^2~b^2=1 is :

3D Geometry PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q44 with Solution

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Correct answer: D

Write the two lines in vector form

First line: $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}=t$ so $x=-1+3t,\quad y=4t,\quad z=5t.$ Hence a point on it is $A(-1,0,0),$ and its direction vector is $\vec d_1=(3,4,5).$

Second line is already given as $\vec r=(p,2,1)+\lambda(2,3,4).$ So a point on it is $B(p,2,1),$ and its direction vector is $\vec d_2=(2,3,4).$
Formula for shortest distance between two skew lines

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

Here $\overrightarrow{AB}=B-A=(p+1,2,1).$
Compute the cross product
$\begin{vmatrix} \hat i & \hat j & \hat k\\ 3&4&5\\ 2&3&4 \end{vmatrix}$$ $$=\hat i(16-15)-\hat j(12-10)+\hat k(9-8)=(1,-2,1).$$ Therefore, $$|\vec d_1\times \vec d_2|=\sqrt{1^2+(-2)^2+1^2}=\sqrt6.$$$
Use the given shortest distance

Given $D=\frac1{\sqrt6}.$

Now
$=(p+1,2,1)\cdot(1,-2,1) =(p+1)-4+1=p-2.$$ So $$D=\frac{|p-2|}{\sqrt6}.$$ Equating with the given value, $$\frac{|p-2|}{\sqrt6}=\frac1{\sqrt6}$$ $$|p-2|=1$$ $$p=1\text{ or }3.$$ Hence $$a=1,\quad b=3\quad (a<b).$$$
Ellipse and its latus rectum

The ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

Since $a=1$ and $b=3$ , this becomes $\frac{x^2}{1^2}+\frac{y^2}{3^2}=1.$

So the semi-major axis is $3$ and semi-minor axis is $1$ .

For an ellipse with semi-major axis $A$ and semi-minor axis $B$ , length of latus rectum is $\frac{2B^2}{A}.$

Here $A=3,\quad B=1.$

Therefore, $\text{latus rectum length}=\frac{2\cdot 1^2}{3}=\frac23.$
Match with options

$\boxed{\frac23}$

So the correct option is D.

More from 3D Geometry

View written solutionFree

Correct answer: D

Write the two lines in vector form

First line: $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}=t$ so $x=-1+3t,\quad y=4t,\quad z=5t.$ Hence a point on it is $A(-1,0,0),$ and its direction vector is $\vec d_1=(3,4,5).$

Second line is already given as $\vec r=(p,2,1)+\lambda(2,3,4).$ So a point on it is $B(p,2,1),$ and its direction vector is $\vec d_2=(2,3,4).$
Formula for shortest distance between two skew lines

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

Here $\overrightarrow{AB}=B-A=(p+1,2,1).$
Compute the cross product
$\begin{vmatrix} \hat i & \hat j & \hat k\\ 3&4&5\\ 2&3&4 \end{vmatrix}$$ $$=\hat i(16-15)-\hat j(12-10)+\hat k(9-8)=(1,-2,1).$$ Therefore, $$|\vec d_1\times \vec d_2|=\sqrt{1^2+(-2)^2+1^2}=\sqrt6.$$$
Use the given shortest distance

Given $D=\frac1{\sqrt6}.$

Now
$=(p+1,2,1)\cdot(1,-2,1) =(p+1)-4+1=p-2.$$ So $$D=\frac{|p-2|}{\sqrt6}.$$ Equating with the given value, $$\frac{|p-2|}{\sqrt6}=\frac1{\sqrt6}$$ $$|p-2|=1$$ $$p=1\text{ or }3.$$ Hence $$a=1,\quad b=3\quad (a<b).$$$
Ellipse and its latus rectum

The ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

Since $a=1$ and $b=3$ , this becomes $\frac{x^2}{1^2}+\frac{y^2}{3^2}=1.$

So the semi-major axis is $3$ and semi-minor axis is $1$ .

For an ellipse with semi-major axis $A$ and semi-minor axis $B$ , length of latus rectum is $\frac{2B^2}{A}.$

Here $A=3,\quad B=1.$

Therefore, $\text{latus rectum length}=\frac{2\cdot 1^2}{3}=\frac23.$
Match with options

$\boxed{\frac23}$

So the correct option is D.

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