1

### JEE Main 2017 (Online) 9th April Morning Slot

If the vector $\overrightarrow b = 3\widehat j + 4\widehat k$ is written as the sum of a vector $\overrightarrow {{b_1}} ,$ paralel to $\overrightarrow a = \widehat i + \widehat j$ and a vector $\overrightarrow {{b_2}} ,$ perpendicular to $\overrightarrow a ,$ then $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}}$ is equal to :
A
$- 3\widehat i + 3\widehat j - 9\widehat k$
B
$6\widehat i - 6\widehat j + {9 \over 2}\widehat k$
C
$- 6\widehat i + 6\widehat j - {9 \over 2}\widehat k$
D
$3\widehat i - 3\widehat j + 9\widehat k$

## Explanation

$\overrightarrow {{b_1}} = {{\left( {\overrightarrow {{b_1}} .\overrightarrow a } \right)\widehat a} \over 1}$

=   $\left\{ {{{\left( {3\widehat j + 4\widehat k} \right).\left( {\widehat i + \widehat j} \right)} \over {\sqrt 2 }}} \right\}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)$

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

$\overrightarrow {{b_1}} + \overrightarrow {{b_2}} = \overrightarrow b$

$\Rightarrow$   $\overrightarrow {{b_2}} = \overrightarrow b - \overrightarrow {{b_1}}$

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

$\Rightarrow$   $\overrightarrow {{b_2}}$ = $- {3 \over 2}\widehat i + {3 \over 2}\widehat j + 4\widehat k$

&  $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{3 \over 2}} & {{3 \over 2}} & 0 \cr { - {3 \over 2}} & {{3 \over 2}} & 4 \cr } } \right|$

$\Rightarrow$   $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \widehat i\left( 6 \right) - \widehat j\left( 6 \right) + \widehat k\left( { - {9 \over 4} + {9 \over 4}} \right)$

$\Rightarrow$   $6\widehat i - 6\widehat j + {9 \over 2}\widehat k$
2

### JEE Main 2018 (Offline)

The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane, x + y + z = 7 is :
A
$\sqrt {{2 \over 3}}$
B
${2 \over {\sqrt 3 }}$
C
${2 \over 3}$
D
${1 \over 3}$

## Explanation PQ is the projection of line segment AB on the plane x + y + z = 7

P and Q are called foot of perpendicular on the plane x + y + z = 7

Let P = (x, y, z) then

${{x - 5} \over 1} = {{y + 1} \over 1} = {{z - 4} \over 1} = {{ - \left( {5 - 1 + 4} \right)} \over {{1^2} + {1^2} + {1^2}}}$

$\Rightarrow \,\,\,\,x - 5 = y + 1 = z - 4 = - {8 \over 3}$

$\therefore\,\,\,$ x = ${7 \over 3}$ , y = $-$ ${{11} \over 3},$ z = ${4 \over 3}$

$\therefore\,\,\,$ Point P = $\left( {{7 \over 3}, - {{11} \over 3},{4 \over 3}} \right)$

Let Q = (x1, y1, z1) then

${{{a_1} - 4} \over 1} = {{{y_1} + 1} \over 1} = {{{z_1} - 3} \over 1} = {{ - \left( {4 - 1 + 3} \right)} \over {{1^2} + {1^2} + {1^2}}}$

$\Rightarrow \,\,\,\,$ x1 $-$ 4 = y1 +1= z1 $-$ 3 = $-$ 2

$\therefore\,\,\,$ x = 2, y = $-$ 3, z = 1

$\therefore\,\,\,$ Point Q = (2, $-$ 3, 1)

Now length of PQ is

$\sqrt {{{\left( {{7 \over 3} - 2} \right)}^2} + {{\left( { - {{11} \over 3} + 3} \right)}^2} + {{\left( {{4 \over 3} - 1} \right)}^2}}$

$= \sqrt {{1 \over 9} + {4 \over 9} + {1 \over 9}}$

$= \sqrt {{6 \over 9}}$

$= \sqrt {{2 \over 3}}$
3

### JEE Main 2018 (Offline)

Let $\overrightarrow u$ be a vector coplanar with the vectors $\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k$ and $\overrightarrow b = \widehat j + \widehat k$. If $\overrightarrow u$ is perpendicular to $\overrightarrow a$ and $\overrightarrow u .\overrightarrow b = 24$, then ${\left| {\overrightarrow u } \right|^2}$ is equal to
A
336
B
315
C
256
D
84

## Explanation

You should know that, when $\overrightarrow u$ is coplanar with $\overrightarrow a$ and $\overrightarrow b$ then we can write $\overrightarrow u = x\overrightarrow a + y\overrightarrow b$

Here, $\overrightarrow u$ is perpendicular with $\overrightarrow a$ then,

$\overrightarrow u .\overrightarrow a = 0$

$\Rightarrow \,\,\,\,\left( {x\,\overrightarrow a + y\overrightarrow b } \right)\,.\overrightarrow a = 0$

$\Rightarrow \,\,\,\,x\,.\overrightarrow {\,a} \,.\,\overrightarrow a \, + \,y\,.\,\overrightarrow a \,.\,\overrightarrow b = 0$

$\Rightarrow \,\,\,\,x\,.\,{\left| {\overrightarrow a } \right|^2} + \,y\overrightarrow a \,.\,\overrightarrow b = 0$

[ As $\left| {\overrightarrow a } \right| = \sqrt {{2^2} + {3^2} + {{\left( { - 1} \right)}^2}} = \sqrt {14}$

and $\overrightarrow a .\overrightarrow b = \left( {2\widehat i + 3\widehat j - \widehat k} \right).\left( {\widehat j + \widehat k} \right)$

$= \,\,\,\,\left( {2.0 + 3.1 + \left( { - 1} \right).1} \right)$

$= \,\,\,\,2$ ]

$\Rightarrow \,\,\,\,x\,.\,\left( {14} \right) + y\,.\,2 = 0$

$\Rightarrow \,\,\,\,7x\, + \,y\, = 0........\left( 1 \right)$

Given, $\overrightarrow u \,.\,\overrightarrow b = 24$

$\Rightarrow \,\,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow b = 24$

$\Rightarrow \,\,\,\,x\left( {\overrightarrow a .\overrightarrow b } \right) + y{\left| {\overrightarrow b } \right|^2} = 24$

$\Rightarrow \,\,\,\,x.2 + y.{\left( {\sqrt {{1^2} + {1^2}} } \right)^2} = 24$

$\Rightarrow \,\,\,\,2x + 2y = 24$

$\Rightarrow \,\,\,\,x + y = 12.......\left( 2 \right)$

By solvig (1) and (2) we get,

x = - 2 and y = 14

Now, ${\left| {\overrightarrow u } \right|^2} = \overrightarrow u .\overrightarrow u$

$= \,\,\,\,\left( {x\overrightarrow a + y\overrightarrow b } \right).\overrightarrow u$

$= \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u$

$= \,\,\,\,x\overrightarrow a .\overrightarrow u + y\overrightarrow b .\overrightarrow u$

$= \,\,\,\,0 + 14 \times 24$ [as $\overrightarrow a .\overrightarrow u = 0$ and $\overrightarrow u .\overrightarrow b = 24]$

$=\,\,\,\,$ 336
4

### JEE Main 2018 (Offline)

If L1 is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L2 is the line of intersection of the planes x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0, then the distance of the origin from the plane, containing the lines L1 and L2, is :
A
${1 \over {\sqrt 2 }}$
B
${1 \over {4\sqrt 2 }}$
C
${1 \over {3\sqrt 2 }}$
D
${1 \over {2\sqrt 2 }}$

## Explanation L1 is the line of intersection of plane 1 and plane 2.

L2 is the line of intersection of plane $3$ and plane 4.

Line L1 and L2 are present on plane 5.

Now we have to find the distance of origin from the plane 5.

The plane 5 is passing through the Line L1,

and L1 is the line of intersection of first two planes.

$\therefore\,\,\,\,$ equation of plane 5,

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

Now find a point on L2 which is line of intersection of

x + 2y $-$ z $-$ 3 = 0 $\,\,\,\,$ .....(1)

and 3x $-$ y + 2z $-$ 1 = 0 $\,\,\,\,$ .....(2)

Put x = 0 at both (1) and (2),

2y $-$ z = 3 $\,\,\,\,.....$(3)

$-$y + 2z = 1 $\,\,\,\,....$ (4)

by solving (3) and (4) we get

$y = {7 \over 3},\,z = {5 \over 3}$

So, the point on line L2 is $= \left( {0,{7 \over 3},{5 \over 3}} \right)$

This point is also present on the plane $5.$

So, putting this point on equation of plane $5,$

$- {{14} \over 5} + 5 - 2 + \lambda \left( { - {7 \over 3} + {5 \over 3}} \right) = 0$

$\Rightarrow \,\,\,\, - {5 \over 3} + {\lambda \over 3} = 0$

$\Rightarrow \,\,\,\,\,\lambda = 3$

$\therefore\,\,\,\,$ equation of plane 5 is

2x $-$ 2y + 3z $-$ 2 + $\lambda$(x $-$ y + z + 1) = 0

$\Rightarrow \,\,\,\,$ 2x $-$ 2y + 3z $-$ 2 + 5 (x $-$ y + z + 1) = 0

$\Rightarrow \,\,\,\,$ 7x $-$ 7y + 8z + 3 = 0

Distance of this plane 5 from (0, 0, 0)

$= \,\,\,\,{3 \over {\sqrt {49 + 49 + 64} }}$

$= \,\,\,\,{3 \over {\sqrt {162} }}$

$= \,\,\,\,{3 \over {9\sqrt 2 }}$

$= \,\,\,\,{1 \over {3\sqrt 2 }}$