1

### JEE Main 2019 (Online) 11th January Morning Slot

The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle ${\pi \over 4}$ with the plane y $-$ z + 5 = 0 are :
A
2, $-$1, 1
B
$2\sqrt 3 ,1, - 1$
C
$\sqrt 2 ,1, - 1$
D
$\sqrt 2 , - \sqrt 2$

## Explanation

Let the equation of plane be

a(x $-$ 0) + b(y + 1) + c(z $-$ 0) = 0

It passes through (0, 0, 1) then

b + c = 0     . . . . (1)

Now cos ${\pi \over 4}$ = ${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {c^2}} }}$

$\Rightarrow$  a2 $=$ $-$ 2bc and b $=$ $-$ c

we get a2 $=$ 2c2

$\Rightarrow$  a $=$ $\pm$ $\sqrt 2$ c

$\Rightarrow$  direction ratio (a, b, c) = ($\sqrt 2$, $-$1, 1) or ($\sqrt 2$, 1, $-$ 1)
2

### JEE Main 2019 (Online) 11th January Evening Slot

Two lines ${{x - 3} \over 1} = {{y + 1} \over 3} = {{z - 6} \over { - 1}}$ and ${{x + 5} \over 7} = {{y - 2} \over { - 6}} = {{z - 3} \over 4}$ intersect at the point R. The reflection of R in the xy-plane has coordinates :
A
(2, 4, 7)
B
(2, $-$ 4, $-$7)
C
(2, $-$ 4, 7)
D
($-$ 2, 4, 7)

## Explanation

Point on L1 ($\lambda$ + 3, 3$\lambda$ $-$ 1, $-$$\lambda$ + 6)

Point on L2 (7$\mu$ $-$ 5, $-$6$\mu$ + 2, 4$\mu$ + 3

$\Rightarrow$  $\lambda$ + 3 = 7$\mu$ $-$ 5      . . . . (i)

3$\lambda$ $-$ 1 = $-$6$\mu$ + 2       . . . .(ii)

$\Rightarrow$  $\lambda$ = $-$1, $\mu$ = 1

point R(2, $-$ 4, 7)

Reflection is (2, $-$4, $-$ 7)
3

### JEE Main 2019 (Online) 11th January Evening Slot

If the point (2, $\alpha$, $\beta$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$\alpha$ – 3$\beta$ is equal to
A
12
B
7
C
17
D
5

## Explanation

Normal vector of plane

$= \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & { - 5} & 0 \cr 4 & { - 4} & 5 \cr } } \right|$

$= - 4\left( {5\widehat i + 2\widehat j - 3\widehat k} \right)$

equation of plane is

5(x $-$ 7) + 2y $-$ 3(z $-$ 6) = 0

5x + 2y $-$ 3z = 17

5 $\times$ 2 + 2$\alpha$ $-$ 3$\beta$ = 17

$\therefore$  2$\alpha$ $-$ 3$\beta$ = 17 $-$ 10 = 7
4

### JEE Main 2019 (Online) 11th January Evening Slot

Let $\sqrt 3 \widehat i + \widehat j,$    $\widehat i + \sqrt 3 \widehat j$  and   $\beta \widehat i + \left( {1 - \beta } \right)\widehat j$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is ${3 \over {\sqrt 2 }}$, then the sum of all possible values of $\beta$ is :
A
4
B
1
C
2
D
3

## Explanation

Angle bisector is x $-$ y = 0

$\Rightarrow$  ${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$

$\Rightarrow$  $\left| {2\beta - 1} \right| = 3$

$\Rightarrow$  $\beta$ = 2 or $-$ 1