1

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

The equation of the plane containing the straight line ${x \over 2} = {y \over 3} = {z \over 4}$ and perpendicular to the plane containing the straight lines ${x \over 3} = {y \over 4} = {z \over 2}$ and ${x \over 4} = {y \over 2} = {z \over 3}$ is :
A
x $-$ 2y + z = 0
B
3x + 2y $-$ 3z = 0
C
x + 2y $-$ 2z = 0
D
5x + 2y $-$ 4z = 0

## Explanation

Vector $\bot$ to given plane = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 4 & 2 \cr 4 & 2 & 3 \cr } } \right|$

= $\widehat i\left( {12 - 4} \right) - \widehat j\left( {9 - 8} \right) + \widehat k\left( {6 - 16} \right)$

= $8\widehat i - \widehat j - 10\widehat k\,$      . . . . (1)

Vector parallel to given line

= $2\widehat i + 3\widehat j + 4\widehat k\,\,\,\,$      . . . (2)

Vector $\bot \,\,\,$ to both (1) & (2) vector

= $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 8 & { - 1} & { - 10} \cr 2 & 3 & 4 \cr } } \right|$

= $\widehat i\left( { - 4 + 30} \right) - \widehat j\left( {32 + 20} \right) + \widehat k\left( {24 + 2} \right)$

= $26\widehat i - 52\widehat j + 26\widehat k$

Dr's of normal of required plane is

(26, $-$52, 26) $\Rightarrow$ (1, $-$2, 1)

Equation of plane whose Dr's of Normal is (1, $-$2, 1) and passes through origin

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

x $-$ 2y + z = 0
2

### 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
3

### 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

$\Rightarrow$  $\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$
4

### 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

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