1

### JEE Main 2017 (Online) 8th April Morning Slot

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\widehat i - 6\widehat j$ and $3\widehat i + 4\widehat j - 12\widehat k,$ is :
A
26
B
65
C
20
D
52

## Explanation

When diagonal ${\overrightarrow {{d_1}} }$ and ${\overrightarrow {{d_2}} }$ are given of a parallelogram then the area of parallelogram = ${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$

Given, ${\overrightarrow {{d_1}} }$ = 8$\widehat i$ $-$ 6$\widehat j$ + 0$\widehat k$

and    ${\overrightarrow {{d_2}} }$ = 3$\widehat i$ + 4$\widehat j$ $-$ 12$\widehat k$

$\therefore\,\,\,$ ${\overrightarrow {{d_1}} }$ $\times$ ${\overrightarrow {{d_2}} }$   =   $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 8 & { - 6} & 0 \cr 3 & 4 & { - 12} \cr } } \right|$

= 72 $\widehat i$ $-$ ($-$ 96) $\widehat j$ + 50$\widehat k$

= 72 $\widehat i$ + 96 $\widehat j$ + 50 $\widehat k$

$\therefore\,\,\,$ $\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$ = $\sqrt {{{72}^2} + {{96}^2} + {{50}^2}}$

= $\sqrt {16900}$

= 130

$\therefore\,\,\,$ Area of parallelogram = ${1 \over 2}\left| {\overrightarrow {{d_1}} \times \overrightarrow {{d_2}} } \right|$ = ${1 \over 2}$ $\times$ 130 = 65
2

### JEE Main 2017 (Online) 8th April Morning Slot

The line of intersection of the planes $\overrightarrow r .\left( {3\widehat i - \widehat j + \widehat k} \right) = 1\,\,$ and
$\overrightarrow r .\left( {\widehat i + 4\widehat j - 2\widehat k} \right) = 2,$ is :
A
${{x - {4 \over 7}} \over { - 2}} = {y \over 7} = {{z - {5 \over 7}} \over {13}}$
B
${{x - {4 \over 7}} \over 2} = {y \over { - 7}} = {{z + {5 \over 7}} \over {13}}$
C
${{x - {6 \over {13}}} \over 2} = {{y - {5 \over {13}}} \over { - 7}} = {z \over { - 13}}$
D
${{x - {6 \over {13}}} \over 2} = {{y - {5 \over {13}}} \over 7} = {z \over { - 13}}$

## Explanation

$\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}}$

$\Rightarrow$   $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 1 \cr 1 & 4 & { - 2} \cr } } \right| = \widehat i\left( { - 2} \right) - \widehat j\left( { - 7} \right) + \widehat k\left( {13} \right)$

$\overrightarrow n = - 2\widehat i + 7\widehat j + 13\widehat k$

Now,

$3x - y + z = 1$

$x + 4y - 2z = 2$

but   z $=$ 0 & solving the given

x $=$ 6/13 & y = 5/13

$\therefore$   required equation of a line is

${{x - 6/13} \over 2} = {{y - 5/13} \over { - 7}} = {z \over { - 13}}$
3

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

If x = a, y = b, z = c is a solution of the system of linear equations

x + 8y + 7z = 0

9x + 2y + 3z = 0

x + y + z = 0

such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals :
A
$-$ 1
B
0
C
1
D
2

## Explanation

Given,

x + 8y + 7z = 0

9x + 2y + 3z = 0

x + y + z = 0

Solving those equations, we get

x = $\lambda$, y = 6$\lambda$, z = -7$\lambda$

This point lies on the plane x + 2y + z = 6

$\therefore$ $\lambda$ + 2(6$\lambda$) + (-7$\lambda$) = 0

$\Rightarrow$ $\lambda$ = 1

$\therefore$ x = 1, y = 6, z = -7

$\therefore$ a = 1, b = 6, c = -7

So, 2a + b + c = 2(1) + 6 + (-7) = 1
4

### 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)

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

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