1

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

An angle between the lines whose direction cosines are gien by the equations,
$l$ + 3m + 5n = 0 and 5$l$m $-$ 2mn + 6n$l$ = 0, is :
A
${\cos ^{ - 1}}\left( {{1 \over 3}} \right)$
B
${\cos ^{ - 1}}\left( {{1 \over 4}} \right)$
C
${\cos ^{ - 1}}\left( {{1 \over 6}} \right)$
D
${\cos ^{ - 1}}\left( {{1 \over 8}} \right)$

## Explanation

Given

l + 3m + 5n = 0

and 5$l$m $-$ 2mn + 6n$l$ = 0

From eq. (1) we have

$l$ = $-$ 3m $-$ 5n

Put the value of $l$ in eq. (2), we get ;

5 ($-$3m $-$5n) m $-$ 2mn + 6n ($-$ 3m $-$ 5n) = 0

$\Rightarrow$  15m2 + 45mn + 30n2 = 0

$\Rightarrow$ m2 + 3mn + 2n2 = 0

$\Rightarrow$  m2 + 2mn + mn + 2n2 = 0

$\Rightarrow$ $\,\,\,$ (m + n) (m + 2n) = 0

$\therefore$  m = $-$ n   or   m = $-$ 2n

For m = $-n,$ $l$ = $-$ 2n

And for m = $-$ 2n, $l$ = n

$\therefore$   ($l$, m, n) = ($-$2n, $-$n, n) Or ($l$, m, n) = (n, $-$ 2n, n)

$\Rightarrow$  ($l$, m, n) = ($-$2, $-$1, 1) Or ($l$, m, n) = (1, $-$ 2, 1)

Therefore, angle between the lines is given as :

cos ($\theta$) = ${{\left( { - 2} \right)\left( 1 \right) + \left( { - 1} \right).\left( { - 2} \right) + \left( 1 \right)\left( 1 \right)} \over {\sqrt 6 .\sqrt 6 }}$

$\Rightarrow$ cos ($\theta$) = ${1 \over 6}$ $\Rightarrow$  $\theta$ =cos$-$1 $\left( {{1 \over 6}} \right)$
2

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

A plane bisects the line segment joining the points (1, 2, 3) and ($-$ 3, 4, 5) at rigt angles. Then this plane also passes through the point :
A
($-$ 3, 2, 1)
B
(3, 2, 1)
C
($-$ 1, 2, 3)
D
(1, 2, $-$ 3)

## Explanation

Since the plane bisects the line joining the points (1, 2, 3) and ($-$3, 4, 5) then the plane passes through the midpoint of the line which is :

$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$  $\equiv$  $\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$ $\equiv$ ($-$1, 3, 4).

As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($-$ 3 $-$ 1, 4 $-$ 2, 5 $-$ 3) = ($-$ 4, 2, 2)

So the equation of the plane is : $-$ 4x + 2y + 2z = $\lambda$

As plane passes through ($-$ 1, 3, 4)

So, $-$ 4($-$ 1) + 2(3) + 2(4) = $\lambda$ $\Rightarrow$ $\lambda$ = 18

Therefore, equation of plane is : $-$ 4x + 2y + 2z = 18

Now, only ($-$ 3, 2, 1) satiesfies the given plane as

$-$ 4($-$ 3) + 2(2) + 2(1) = 18
3

### 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)
4

### 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 $= - 8 + {8 \over 3} + {8 \over 6} = - 4$