1

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

The plane containing the line ${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$ and also containing its projection on the plane 2x + 3y $-$ z = 5, contains which one of the following points ?
A
($-$ 2, 2, 2)
B
(2, 2, 0)
C
(2, 0, $-$ 2)
D
(0, $-$ 2, 2)

## Explanation

The normal vector of required plane

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

$= - 8\widehat i + 8\widehat j + 8\widehat k$

So, direction ratio of normal is $\left( { - 1,1,1} \right)$

So required plane is

$- \left( {x - 3} \right) + \left( {y + 2} \right) + \left( {z - 1} \right) = 0$

$\Rightarrow - x + y + z + 4 = 0$

Which is satisfied by $\left( {2,0, - 2} \right)$
2

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

Let  $\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$ $\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$ and $\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$ be coplanar vectors. Then the non-zero vector $\overrightarrow a \times \overrightarrow c$ is :
A
$- 10\widehat i - 5\widehat j$
B
$- 10\widehat i + 5\widehat j$
C
$- 14\widehat i + 5\widehat j$
D
$- 14\widehat i - 5\widehat j$

## Explanation

$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0$

$\Rightarrow \left| {\matrix{ 1 & 2 & 4 \cr 1 & \lambda & 4 \cr 2 & 4 & {{\lambda ^2} - 1} \cr } } \right| = 0$

$\Rightarrow {\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0$

$\Rightarrow {\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0$

$\Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0$

$\Rightarrow \lambda = 2,3, - 3$

So,   $\lambda$ = 2 (as   $\overrightarrow a$ is parallel to $\overrightarrow c$ for $\lambda$ = $\pm$3)

Hence   $\overrightarrow a \times \overrightarrow c = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 4 \cr 2 & 4 & 3 \cr } } \right|$

$= - 10\widehat i + 5\widehat j$
3

### 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)
4

### 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)