1

### JEE Main 2018 (Online) 15th April Evening Slot

If the position vectors of the vertices A, B and C of a $\Delta$ ABC are respectively $4\widehat i + 7\widehat j + 8\widehat k,$    $2\widehat i + 3\widehat j + 4\widehat k,$     and     $2\widehat i + 5\widehat j + 7\widehat k,$ then the position vectors of the point, where the bisector of $\angle$A meets BC is :
A
${1 \over 2}\left( {4\widehat i + 8\widehat j + 11\widehat k} \right)$
B
${1 \over 3}\left( {6\widehat i + 11\widehat j + 15\widehat k} \right)$
C
${1 \over 3}\left( {6\widehat i + 13\widehat j + 18\widehat k} \right)$
D
${1 \over 4}\left( {8\widehat i + 14\widehat j + 19\widehat k} \right)$

## Explanation

Suppose angular bisector of A meets BC at D(x, , z)

Using angular bisector theorem,

${{AB} \over {AC}}$ = ${{BD} \over {DC}}$

${{BD} \over {DC}}$ = ${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$

= ${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$ = ${6 \over 3}$ = 2

So, D(x, y, z) $\equiv$ $\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$

$\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$

D(x, y, z) $\equiv$ $\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$

Therefore, position vector of point p = ${1 \over 3}$ (6i + 13j + 18k)
2

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

The sum of the intercepts on the coordinate axes of the plane passing through the point ($-$2, $-2,$ 2) and containing the line joining the points (1, $-$1, 2) and (1, 1, 1) is :
A
4
B
$-$ 4
C
$-$ 8
D
12

## Explanation

Equation of plane passing through three given points is :

$\left| {\matrix{ {x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr } } \right| = 0$

$\Rightarrow$  $\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr {1 + 2} & { - 1 + 2} & {2 - 2} \cr {1 + 2} & {1 + 2} & {1 - 2} \cr } } \right| = 0$

$\Rightarrow$  $\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr 3 & 1 & 0 \cr 3 & {30} & { - 1} \cr } } \right| = 0$

$\Rightarrow$  $- x + 3y + 6z - 8 = 0$

$\Rightarrow$  ${x \over 8} - {{3y} \over 8} - {{6z} \over 8} + {8 \over 8} = 0$

$\Rightarrow$  ${x \over 8} - {y \over {{8 \over 3}}} - {z \over {{8 \over 6}}} = - 1$

$\Rightarrow$  ${x \over { - 8}} + {y \over {{8 \over 3}}} + {z \over {{8 \over 6}}} = 1$

$\therefore$   Sum of intercepts $= - 8 + {8 \over 3} + {8 \over 6} = - 4$
3

### 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 + ${{{P^2}} \over {49}}$

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

### 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

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

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