JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 3)
Let the position vectors of two points P and Q be 3$$\widehat i$$ $$-$$ $$\widehat j$$ + 2$$\widehat k$$ and $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 4$$\widehat k$$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, $$-$$1, 2) and ($$-$$2, 1, $$-$$2), respectively. Let lines PR and QS intersect at T. If the vector $$\overrightarrow {TA} $$ is perpendicular to both $$\overrightarrow {PR} $$ and $$\overrightarrow {QS} $$ and the length of vector $$\overrightarrow {TA} $$ is $$\sqrt 5 $$ units, then the modulus of a position vector of A is :
$$\sqrt {171} $$
$$\sqrt {227} $$
$$\sqrt {482} $$
$$\sqrt {5} $$
Explanation
$$\overrightarrow p = 3\widehat i - \widehat j + 2\widehat k$$ & $$\overrightarrow Q = \widehat i + 2\widehat j - 4\widehat k$$
$${\overrightarrow v _{PR}} = (4, - 1,2)$$ & $${\overrightarrow v _{QS}} = ( - 2,1, - 2)$$
_16th_March_Morning_Shift_en_3_1.png)
$${L_{PR}}:\overrightarrow r = (3\widehat i - \widehat j + 2\widehat k) + \lambda (4, - 1,2)$$
$${L_{QS}}:\overrightarrow r = (\widehat i + 2\widehat j - 4\widehat k) + \mu ( - 2,1, - 2)$$
Now T on PR = $$\left\langle {3 + 4\lambda , - 1 - \lambda ,2 + 2\lambda } \right\rangle $$
Similarly T on QS = (1 $$-$$ 2$$\mu$$, 2 + $$\mu$$, $$-$$4 $$-$$ 2$$\mu$$)
For $$\lambda$$ & $$\mu$$ : $$\left. \matrix{ 3 + 4\lambda = 1 - 2\mu \Rightarrow \mu + 2\lambda = - 1 \hfill \cr - 1 - \lambda = 2 + \mu \Rightarrow \mu + \lambda = - 3 \hfill \cr} \right\}\matrix{ {\lambda = 2} \cr {\mu = - 5} \cr } $$
$$ \Rightarrow $$ T : (11, $$-$$3, 6)
D.R. of TA = $${\overrightarrow v _{QS}}$$ $$\times$$ $${\overrightarrow v _{PR}}$$
= $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 2} & 1 & { - 2} \cr 4 & { - 1} & 2 \cr } } \right| = 0\widehat i - 4\widehat j - 2\widehat k$$
$${L_{TA}}:\overrightarrow r = \left( {11\widehat i - 3\widehat j + 6\widehat k} \right) + \lambda ( - 4\widehat j - 2\widehat k)$$
Now A = (11, $$-$$3 $$-$$4$$\lambda$$, 6 $$-$$ 2$$\lambda$$)
Given, TA = $$\sqrt 5 $$
$${( - 3 + 4\lambda + 3)^2} + {(6 + 2\lambda - 6)^2} = 5$$
$$16{\lambda ^2} + 4{\lambda ^2} = 5$$
20$$\lambda$$2 = 5
$$\lambda$$ = $$\pm$$$${1 \over 2}$$
$$\matrix{ {A:(11, - 3 - 2,6 - 1)} & ; & {A:(11, - 3 + 2,6 + 1)} \cr {|A|\, = \sqrt {121 + 25 + 25} } & ; & {|A|\, = \sqrt {121 + 1 + 49} } \cr { = \sqrt {171} } & ; & {\sqrt {171} } \cr } $$
$${\overrightarrow v _{PR}} = (4, - 1,2)$$ & $${\overrightarrow v _{QS}} = ( - 2,1, - 2)$$
$${L_{PR}}:\overrightarrow r = (3\widehat i - \widehat j + 2\widehat k) + \lambda (4, - 1,2)$$
$${L_{QS}}:\overrightarrow r = (\widehat i + 2\widehat j - 4\widehat k) + \mu ( - 2,1, - 2)$$
Now T on PR = $$\left\langle {3 + 4\lambda , - 1 - \lambda ,2 + 2\lambda } \right\rangle $$
Similarly T on QS = (1 $$-$$ 2$$\mu$$, 2 + $$\mu$$, $$-$$4 $$-$$ 2$$\mu$$)
For $$\lambda$$ & $$\mu$$ : $$\left. \matrix{ 3 + 4\lambda = 1 - 2\mu \Rightarrow \mu + 2\lambda = - 1 \hfill \cr - 1 - \lambda = 2 + \mu \Rightarrow \mu + \lambda = - 3 \hfill \cr} \right\}\matrix{ {\lambda = 2} \cr {\mu = - 5} \cr } $$
$$ \Rightarrow $$ T : (11, $$-$$3, 6)
D.R. of TA = $${\overrightarrow v _{QS}}$$ $$\times$$ $${\overrightarrow v _{PR}}$$
= $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 2} & 1 & { - 2} \cr 4 & { - 1} & 2 \cr } } \right| = 0\widehat i - 4\widehat j - 2\widehat k$$
$${L_{TA}}:\overrightarrow r = \left( {11\widehat i - 3\widehat j + 6\widehat k} \right) + \lambda ( - 4\widehat j - 2\widehat k)$$
Now A = (11, $$-$$3 $$-$$4$$\lambda$$, 6 $$-$$ 2$$\lambda$$)
Given, TA = $$\sqrt 5 $$
$${( - 3 + 4\lambda + 3)^2} + {(6 + 2\lambda - 6)^2} = 5$$
$$16{\lambda ^2} + 4{\lambda ^2} = 5$$
20$$\lambda$$2 = 5
$$\lambda$$ = $$\pm$$$${1 \over 2}$$
$$\matrix{ {A:(11, - 3 - 2,6 - 1)} & ; & {A:(11, - 3 + 2,6 + 1)} \cr {|A|\, = \sqrt {121 + 25 + 25} } & ; & {|A|\, = \sqrt {121 + 1 + 49} } \cr { = \sqrt {171} } & ; & {\sqrt {171} } \cr } $$
Comments (0)
