1

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

If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of $\Delta$ABC is :
A
${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 1$
B
${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 3$
C
${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = {1 \over 9}$
D
${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 9$

## Explanation

Suppose centroid be (h, k, $l$)

$\therefore$   x $-$ intp $=$ 3h, y $-$ intp $=$ 3k, z $-$ intp $=$ 3$l$

Equation ${x \over {3h}} + {y \over {3k}} + {z \over {3l}} = 1$

$\therefore$   Distance from (0, 0, 0)

$\left| {{{ - 1} \over {\sqrt {{1 \over {9{h^2}}} + {1 \over {9{k^2}}} + {1 \over {9{l^2}}}} }}} \right| = 3$

$\Rightarrow$   ${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 1$
2

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

If the line, ${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}}$ lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, ${{x - 1} \over {12}} = {y \over 9} = {z \over 4}$ is :
A
2
B
1
C
0
D
3

## Explanation

Point (3, $-$ 2, $-$ $\lambda$) on p line 2x $-$ 4y + 3z $-$ 2 $=$ 0

$=$ 6 + 8 $-$ 3$\lambda$ $-$ 2 = 0

$=$ 3$\lambda$ $=$ 12

$\lambda$ $=$ 4

Now,

${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = {k_1}$          . . .(i)

${{x - 1} \over {12}} = {y \over 9} = {z \over 4} = {k_2}$                 . . .(ii)

Point on equation (i) P (k1 + 3, $-$ k1 $-$ 2, $-$ 2k1 $-$ 4)

Point on equation (ii) Q(12k2 + 1, 9k2, 4k2)

k1 + 3 $=$ 12k2 + 1 $\left| { - {k_1} - 2 = 9{k_2}} \right|$ $-$ 2k1 $-$ 4 $=$ 4k2

k2  $=$  0

k1   $=$  $-$ 2

p (1, 0, 0) lie on equation of a line 1

gives shortest distance $=$ 0
3

### 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

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

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

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

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

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

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

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

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

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

$\Rightarrow$   $6\widehat i - 6\widehat j + {9 \over 2}\widehat k$
4

### 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

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

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

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

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