MCQ
Let the values of p, for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{ r }=( p \hat{ i }+2 \hat{ j }+\hat{ k })+\lambda(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })$ is $\frac{1}{\sqrt{6}}$, be a, b, $( a < b )$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is :-
  • A
    9
  • B
    $\frac{3}{2}$
  • $\frac{2}{3}$
  • D
    18

Answer

Correct option: C.
$\frac{2}{3}$
(C) $\frac{2}{3}$
$\text { shortest distance }=\frac{|(\overline{a}-\overline{b})| \cdot(\overline{p} \times \overline{q})}{|\overline{p} \times \overline{q}|}$
where
$\overline{a}=-\hat{i}+0 \hat{j}+0 \hat{k}$
$\quad$$\quad$$\quad$$>\overline{ a }-\overline{ b }=(-1- p ) \hat{ i }-2 \hat{ j }-\hat{ k }$
$\overline{ b }= p \hat{ i }+2 \hat{ j }+\hat{ k } $
$\overline{ p }=3 \hat{ i }+4 \hat{ j }+5 \hat{ k } $
$\overline{ q }=2 \hat{ i }+3 \hat{ j }+4 \hat{ k } $
$\frac{1}{16}=\frac{|-1- p +4-1|}{\sqrt{6}} $
$|- p +2|=1 $
$p =3 \quad \& \quad q =1 $
$\frac{ x ^2}{1^2}+\frac{ y ^2}{3^3}=1$
$L \cdot R =\frac{2 a ^2}{b}=\frac{2 \times 1}{3}=\frac{2}{3}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $2\sec 2\alpha = \tan \beta + \cot \beta ,$ then one of the values of $\alpha + \beta $ is
The solution curve, of the differential equation $2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}+3=5 \frac{\mathrm{dy}}{\mathrm{dx}}$, passing through the point $(0,1)$ is a conic, whose vertex lies on the line :
If the function $\mathrm{f}$ defined on $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by $f(x)=\left\{\begin{array}{ll}{\frac{1}{x} \log _{e}\left(\frac{1+3 x}{1-2 x}\right)} & {, \text { when } x \neq 0} \\ {k} & {, \text { when } x=0}\end{array}\right.$ is continuous, then $\mathrm{k}$ is equal to
Ten students are seated at random in a row. The probability that two particular students are not seated side by side is
Let $\alpha$ and $\beta$ be the roots of $x^2-x-1=0$, with $\alpha>\beta$. For all positive integers $n$, define

$a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}, n \geq 1$

$b_1=1 \text { and } b_n=a_{n-1}+a_{n+1}, n \geq 2.$

Then which of the following options is/are correct?

$(1)$ $a_1+a_2+a_3+\ldots . .+a_n=a_{n+2}-1$ for all $n \geq 1$

$(2)$ $\sum_{n=1}^{\infty} \frac{ a _{ n }}{10^{ n }}=\frac{10}{89}$

$(3)$ $\sum_{n=1}^{\infty} \frac{b_n}{10^n}=\frac{8}{89}$

$(4)$ $b=\alpha^n+\beta^n$ for all $n>1$

An ellipse is drawn by taking a diameter of the circle ${\left( {x - 1} \right)^2} + {y^2} = 1$ as its semi-minor axis and a diameter of the circle ${x^2} + {\left( {y - 2} \right)^2} = 4$ is semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :
$\int_0^{\pi /2} {{{\sin }^5}x\,dx = } $
If $A = \left[ {\begin{array}{*{20}{c}}0&1&{ - 2}\\{ - 1}&0&5\\2&{ - 5}&0\end{array}} \right]$, then
For a sequence $ < {a_n} > ,\;{a_1} = 2$ and $\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{1}{3}$. Then $\sum\limits_{r = 1}^{20} {{a_r}} $ is
Let $f _1: R \rightarrow R , f _2:[0, \infty) \rightarrow R , f _3: R \rightarrow R$ and $f _4: R \rightarrow[0, \infty)$ be defined by

$f_1(x)=\left\{\begin{array}{lll}|x| & \text { if } & x<0, \\ e^x & \text { if } & x \geq 0 ;\end{array}\right.$

$f_2(x)=x^2$

$f_3(x)=\left\{\begin{array}{ccc}\sin x & \text { if } & x < 0, \\ x & \text { if } & x \geq 0\end{array}\right.$ and

$f_4(x)=\left\{\begin{array}{ccc}f_2\left(f_1(x)\right) & \text { if } & x < 0, \\ f_2\left(f_1(x)\right)-1 & \text { if } & x \geq 0\end{array}\right.$

List $I$ List $II$
$P.$ $ f_4$ is $1.$ onto but not one-one
$Q.$ $f_3$ is $2.$ neither continuous nor one-one
$R.$ $f _2 \circ f _1$ is $3.$ differentiable but not one-one
$S.$ $ f_2$ is $4.$ continuous and one-one

Codes: $ \quad P \quad Q \quad R \quad S $