1

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

If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :
A
ab'  +  bc'  +  1  =  0
B
cc'  +  a   +  a'  =  0
C
bb'  +  cc'  +  1  =  0
D
aa'  +  c  +  c'  =  0

## Explanation

Equation of 1st line is

${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$

Dr's of 1st line = ($a$, 1 , c)

Equation of 2nd line is

${{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}$

Dr's of 2nd line = ($a'$, c' , 1)

Lines are perpendicular, so the dot product of the Dr's of two lines are zero.

$\therefore$ $aa$' + c + c' = 0
2

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

Let  $\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$   $\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$,    $\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$   be three vectors such that the projection vector of $\overrightarrow b$ on $\overrightarrow a$ is $\overrightarrow a$.
If   $\overrightarrow a + \overrightarrow b$   is perpendicular to $\overrightarrow c$ , then $\left| {\overrightarrow b } \right|$ is equal to :
A
$\sqrt {32}$
B
6
C
$\sqrt {22}$
D
4

## Explanation

Projection of $\overrightarrow b$ on $\overrightarrow a$ is $\overrightarrow a$

$\therefore$   ${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$

$\Rightarrow$  ${{{b_1} + {b_2} + 2} \over 2} = 2$

$\Rightarrow$  ${b_1} + {b_2} = 2\,\,\,\,\,\,\,\,\,\,\,....(1)$

and $\overrightarrow a$ + $\overrightarrow b$ is perpendicular to $\overrightarrow c$

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

$\Rightarrow$  $5\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + \sqrt 2 \left( {2\sqrt 2 } \right) = 0$

$\Rightarrow$  $5{b_1} + {b_2} + 10 = 0\,\,\,\,\,\,\,\,\,\,......(2)$

solving (1) & (2)

b1 $=$ $-$ 3 and b2 $=$ 5

$\Rightarrow$  $\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$
3

### JEE Main 2019 (Online) 10th January Morning Slot

Let $\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k,\,\,$   $\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k,$  and  $\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$  be three vectors such that $\overrightarrow b = 2\overrightarrow a$ and $\overrightarrow a$ is perpendicular to $\overrightarrow c$. Then a possible value of $\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right)$ is -
A
(1, 5, 1)
B
(1, 3, 1)
C
$\left( { - {1 \over 2},4,0} \right)$
D
$\left( {{1 \over 2},4, - 2} \right)$

## Explanation

Given $\overrightarrow b = 2\overrightarrow a$

$\therefore$ $4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$

$\Rightarrow 3 - {\lambda _2} = 2{\lambda _1} \Rightarrow 2{\lambda _1} + {\lambda _2} = 3\,\,...(1)$

Given $\overrightarrow a$ is perpendicular to $\overrightarrow c$

$\therefore$ $\overrightarrow a .\overrightarrow c = 0$

$\Rightarrow 6 + 6{\lambda _1} + 3\left( {{\lambda _3} - 1} \right) = 0$

$\Rightarrow 2{\lambda _1} + {\lambda _3} = - 1\,\,\,\,\,\,\,\,\,\,...(2)$

Now $\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right) = \left( {{\lambda _1},3 - 2{\lambda _1}, - 1 - 2{\lambda _1}} \right)$

By checking each option you can see,

when ${\lambda _1}$ = $- {1 \over 2}$

then ${\lambda _2}$ = $3 - 2{\lambda _1}$ = 3 + 1 = 4

and ${\lambda _3}$ = $-1 - 2{\lambda _1}$ = - 1 + 1 = 0
4

### JEE Main 2019 (Online) 10th January Morning Slot

The plane passing through the point (4, –1, 2) and parallel to the lines  ${{x + 2} \over 3} = {{y - 2} \over { - 1}} = {{z + 1} \over 2}$  and  ${{x - 2} \over 1} = {{y - 3} \over 2} = {{z - 4} \over 3}$ also passes through the point -
A
(1, 1, $-$ 1)
B
(1, 1, 1)
C
($-$ 1, $-$ 1, $-$1)
D
($-$ 1, $-$ 1, 1)

## Explanation

Let $\overrightarrow n$ be the normal vector to the plane passing through (4, $-$1, 2) and parallel to the lines L1 & L2

then $\overrightarrow n$ = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 2 \cr 1 & 2 & 3 \cr } } \right|$

$\therefore$  $\overrightarrow n$ = $- 7\widehat i - 7\widehat j + 7\widehat k$

$\therefore$  Equation of plane is

$-$ 1(x $-$ 4) $-$ 1(y + 1) + 1(z $-$ 2) = 0

$\therefore$  x + y $-$ z $-$ 1 = 0

Now check options

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