1

### JEE Main 2018 (Online) 16th April Morning Slot

If the angle between the lines, ${x \over 2} = {y \over 2} = {z \over 1}$

and ${{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,$ is ${\cos ^{ - 1}}\left( {{2 \over 3}} \right),$ then p is equal to :
A
${7 \over 2}$
B
${2 \over 7}$
C
$-$ ${7 \over 4}$
D
$-$ ${4 \over 7}$

## Explanation

Let $\theta$ be the angle between the two lines

Here direction cosines of ${x \over 2}$ = ${y \over 2}$ = ${z \over 1}$ are 2, 2, 1

Also second line can be written as :

${{x - 5} \over 2} = {{y - 2} \over {{P \over 7}}} = {{z - 3} \over 4}$

$\therefore$  its direction cosines are 2, ${{P \over 7}}$, 4

Also, cos$\theta$ = ${2 \over 3}$ (Given)

$\because$ cos$\theta$ $= \left| {{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \over {\sqrt {a_1^2 + b_1^2 + c_1^2\sqrt {a_2^2 + b_2^2 + c_2^2} } }}} \right|$

$\Rightarrow$   ${2 \over 3}$ $= \left| {{{\left( {2 \times 2} \right) + \left( {2 \times {P \over 7}} \right) + \left( {1 \times 4} \right)} \over {\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}} \right|$

$= {{4 + {{2P} \over 7} + 4} \over {3 \times \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}$

$\Rightarrow$ ${\left( {4 + {P \over 7}} \right)^2} = 20 + {{{P^2}} \over {49}}$

$\Rightarrow$  16 + ${{8P} \over 7} + {{{P^2}} \over {49}}$ = 20 + ${{{P^2}} \over {49}}$

$\Rightarrow$ ${{8P} \over 7} = 4$ $\Rightarrow$ $P = {7 \over 2}$
2

### JEE Main 2018 (Online) 16th April Morning Slot

Let $\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k$ and a vector $\overrightarrow b$ be such that $\overrightarrow a \times \overrightarrow b = \overrightarrow c$ and $\overrightarrow a .\overrightarrow b = 3.$ Then $\left| {\overrightarrow b } \right|$ equals :
A
${{11} \over 3}$
B
${{11} \over {\sqrt 3 }}$
C
$\sqrt {{{11} \over 3}}$
D
${{\sqrt {11} } \over 3}$

## Explanation

$\because$ $\overrightarrow a$ $=$ $\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3$

&   $\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2$

Now, $\overrightarrow a$ $\times$ $\overrightarrow b$ = $\overrightarrow c$     (Given)

$\Rightarrow$  $\left| {\vec a} \right|\left| {\vec b} \right|\sin \theta = \left| {\overrightarrow c } \right|$

$\Rightarrow$  $\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \sqrt 2$

also  $\overrightarrow a .\overrightarrow b = 3$

$\Rightarrow$  $\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 3$

Dividing [i] by [iii], we get

tan$\theta$ = ${{\sqrt 2 } \over 3}$

$\therefore$ sin$\theta$ $=$ ${{\sqrt 2 } \over {\sqrt {11} }}$

Substituting value of sin$\theta$ in [i] we get

$\sqrt 3 \left| {\overrightarrow b } \right|{{\sqrt 2 } \over {\sqrt {11} }} = \sqrt 2$

$\left| {\overrightarrow b } \right| = {{\sqrt {11} } \over {\sqrt 3 }}$
3

### JEE Main 2019 (Online) 9th January Morning Slot

Let $\overrightarrow a$ = $\widehat i - \widehat j$, $\overrightarrow b$ = $\widehat i + \widehat j + \widehat k$ and $\overrightarrow c$

be a vector such that $\overrightarrow a$ × $\overrightarrow c$ + $\overrightarrow b$ = $\overrightarrow 0$

and $\overrightarrow a$.$\overrightarrow c$ = 4, then |$\overrightarrow C$|2 is equal to :
A
8
B
$19 \over 2$
C
9
D
$17 \over 2$

## Explanation

Given that,

$\overrightarrow a \times \overrightarrow c + \overrightarrow b = \overrightarrow 0$

$\Rightarrow$  $\overrightarrow a \times \left( {\overrightarrow a \times \overrightarrow c } \right) + \overrightarrow a \times b = \overrightarrow 0$

$\Rightarrow$  $\left( {\overrightarrow a \cdot \overrightarrow c } \right)\overrightarrow a - \left( {\overrightarrow a \cdot \overrightarrow a } \right)\overrightarrow c + \overrightarrow a \times \overrightarrow b = \overrightarrow 0$

given that

$\overrightarrow a \cdot \overrightarrow c = 4$

and $\overrightarrow a \cdot \overrightarrow a = {\left| {\overrightarrow a } \right|^2} = {\left( {\sqrt 2 } \right)^2} = 2$

$\Rightarrow$  $4\overrightarrow a - 2\overrightarrow c + \overrightarrow a \times \overrightarrow b = 0$

Now  $\overrightarrow a \times \overrightarrow b$

$= \left| {\matrix{ i & j & k \cr 1 & { - 1} & 0 \cr 1 & 1 & 1 \cr } } \right|$

$= - \widehat i - \widehat j + 2\widehat k$

$\therefore$  $2\overrightarrow c = 4\left( {\widehat i - \widehat j} \right) + \left( { - \widehat i - \widehat j + \widehat k} \right)$

$= 4\widehat i - 4\widehat j - \widehat i - \widehat j + \widehat k$

$= 3\widehat i - 5\widehat j + \widehat k$

$\therefore$  $\overrightarrow c = {3 \over 2}\widehat i - {5 \over 2}\widehat j + \widehat k$

$\therefore$  $\left| {\overrightarrow c } \right| = \sqrt {{9 \over 4} + {{25} \over 4} + 1}$

$= \sqrt {{{38} \over 4}}$

$= \sqrt {{{19} \over 2}}$

$\therefore$  ${\left| {\overrightarrow c } \right|^2} = {{19} \over 2}$
4

### JEE Main 2019 (Online) 9th January Morning Slot

The equation of the line passing through (–4, 3, 1), parallel
to the plane x + 2y – z – 5 = 0 and intersecting
the line ${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$ is
A
${{x + 4} \over 3} = {{y - 3} \over {-1}} = {{z - 1} \over 1}$
B
${{x + 4} \over 1} = {{y - 3} \over {1}} = {{z - 1} \over 3}$
C
${{x + 4} \over -1} = {{y - 3} \over {1}} = {{z - 1} \over 1}$
D
${{x - 4} \over 2} = {{y + 3} \over {1}} = {{z + 1} \over 4}$

## Explanation

The line L is parallel to the plane P and intersect with line 4 at point R.

Let the coordinate of point R is, (x1, y1, z1) and it passes through L2.

${{{x_1} + 1} \over { - 3}} = {{{y_1} - 3} \over 2} = {{{z_1} - 2} \over { - 1}} = t$

$\therefore$   x1 = $-$1 $-$ 3t, y1 = 3 + 2t, z1 = 2 $-$ t

$\overrightarrow {AR} = \left( {3 - 3t} \right)\widehat i + \left( {2t} \right)\widehat j + \left( {1 - t} \right)\widehat k$

$\overrightarrow n = \widehat i + 2\widehat j - \widehat k$

As $\overrightarrow {AR}$ and $\overrightarrow n$ are perpendicular to each other, So

$\overrightarrow {AR}$ $\cdot$ $\overrightarrow n$ = 0

$\Rightarrow$  (3 $-$ 3t) 1 + (2t)2 + (1 $-$ t) ($-$ 1) = 0

$\Rightarrow$  3 $-$ 3t + 4t $-$ 1 + t = 0

$\Rightarrow$  2 + 2t = 0

$\Rightarrow$  t = $-$ 1

$\therefore$   point R = (2, 1, 3)

$\therefore$  DR of line L is

= (2 $-$ ($-$ 4), $1$ $-$ 3, 3 $-$ 1)

= (6, $-$ 2, 2)

$\therefore$  Equation of line is

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

or ${{x + 4} \over 3} = {{y - 3} \over { - 1}} = {{z - 1} \over 1}$