1

JEE Main 2018 (Online) 16th April Morning Slot

The sum of the intercepts on the coordinate axes of the plane passing through the point ($-$2, $-2,$ 2) and containing the line joining the points (1, $-$1, 2) and (1, 1, 1) is :
A
4
B
$-$ 4
C
$-$ 8
D
12

Explanation

Equation of plane passing through three given points is :

$\left| {\matrix{ {x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr } } \right| = 0$

$\Rightarrow$  $\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr {1 + 2} & { - 1 + 2} & {2 - 2} \cr {1 + 2} & {1 + 2} & {1 - 2} \cr } } \right| = 0$

$\Rightarrow$  $\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr 3 & 1 & 0 \cr 3 & {30} & { - 1} \cr } } \right| = 0$

$\Rightarrow$  $- x + 3y + 6z - 8 = 0$

$\Rightarrow$  ${x \over 8} - {{3y} \over 8} - {{6z} \over 8} + {8 \over 8} = 0$

$\Rightarrow$  ${x \over 8} - {y \over {{8 \over 3}}} - {z \over {{8 \over 6}}} = - 1$

$\Rightarrow$  ${x \over { - 8}} + {y \over {{8 \over 3}}} + {z \over {{8 \over 6}}} = 1$

$\therefore$   Sum of intercepts $= - 8 + {8 \over 3} + {8 \over 6} = - 4$
2

JEE Main 2018 (Online) 16th April Morning Slot

Let $\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k$ and a vector $\overrightarrow b$ be such that $\overrightarrow a \times \overrightarrow b = \overrightarrow c$ and $\overrightarrow a .\overrightarrow b = 3.$ Then $\left| {\overrightarrow b } \right|$ equals :
A
${{11} \over 3}$
B
${{11} \over {\sqrt 3 }}$
C
$\sqrt {{{11} \over 3}}$
D
${{\sqrt {11} } \over 3}$

Explanation

$\because$ $\overrightarrow a$ $=$ $\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3$

&   $\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2$

Now, $\overrightarrow a$ $\times$ $\overrightarrow b$ = $\overrightarrow c$     (Given)

$\Rightarrow$  $\left| {\vec a} \right|\left| {\vec b} \right|\sin \theta = \left| {\overrightarrow c } \right|$

$\Rightarrow$  $\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \sqrt 2$

also  $\overrightarrow a .\overrightarrow b = 3$

$\Rightarrow$  $\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 3$

Dividing [i] by [iii], we get

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

$\therefore$ sin$\theta$ $=$ ${{\sqrt 2 } \over {\sqrt {11} }}$

Substituting value of sin$\theta$ in [i] we get

$\sqrt 3 \left| {\overrightarrow b } \right|{{\sqrt 2 } \over {\sqrt {11} }} = \sqrt 2$

$\left| {\overrightarrow b } \right| = {{\sqrt {11} } \over {\sqrt 3 }}$
3

JEE Main 2019 (Online) 9th January Morning Slot

Let $\overrightarrow a$ = $\widehat i - \widehat j$, $\overrightarrow b$ = $\widehat i + \widehat j + \widehat k$ and $\overrightarrow c$

be a vector such that $\overrightarrow a$ × $\overrightarrow c$ + $\overrightarrow b$ = $\overrightarrow 0$

and $\overrightarrow a$.$\overrightarrow c$ = 4, then |$\overrightarrow C$|2 is equal to :
A
8
B
$19 \over 2$
C
9
D
$17 \over 2$

Explanation

Given that,

$\overrightarrow a \times \overrightarrow c + \overrightarrow b = \overrightarrow 0$

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

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

given that

$\overrightarrow a \cdot \overrightarrow c = 4$

and $\overrightarrow a \cdot \overrightarrow a = {\left| {\overrightarrow a } \right|^2} = {\left( {\sqrt 2 } \right)^2} = 2$

$\Rightarrow$  $4\overrightarrow a - 2\overrightarrow c + \overrightarrow a \times \overrightarrow b = 0$

Now  $\overrightarrow a \times \overrightarrow b$

$= \left| {\matrix{ i & j & k \cr 1 & { - 1} & 0 \cr 1 & 1 & 1 \cr } } \right|$

$= - \widehat i - \widehat j + 2\widehat k$

$\therefore$  $2\overrightarrow c = 4\left( {\widehat i - \widehat j} \right) + \left( { - \widehat i - \widehat j + \widehat k} \right)$

$= 4\widehat i - 4\widehat j - \widehat i - \widehat j + \widehat k$

$= 3\widehat i - 5\widehat j + \widehat k$

$\therefore$  $\overrightarrow c = {3 \over 2}\widehat i - {5 \over 2}\widehat j + \widehat k$

$\therefore$  $\left| {\overrightarrow c } \right| = \sqrt {{9 \over 4} + {{25} \over 4} + 1}$

$= \sqrt {{{38} \over 4}}$

$= \sqrt {{{19} \over 2}}$

$\therefore$  ${\left| {\overrightarrow c } \right|^2} = {{19} \over 2}$
4

JEE Main 2019 (Online) 9th January Morning Slot

The equation of the line passing through (–4, 3, 1), parallel
to the plane x + 2y – z – 5 = 0 and intersecting
the line ${{x + 1} \over { - 3}} = {{y - 3} \over 2} = {{z - 2} \over { - 1}}$ is
A
${{x + 4} \over 3} = {{y - 3} \over {-1}} = {{z - 1} \over 1}$
B
${{x + 4} \over 1} = {{y - 3} \over {1}} = {{z - 1} \over 3}$
C
${{x + 4} \over -1} = {{y - 3} \over {1}} = {{z - 1} \over 1}$
D
${{x - 4} \over 2} = {{y + 3} \over {1}} = {{z + 1} \over 4}$

Explanation

The line L is parallel to the plane P and intersect with line 4 at point R.

Let the coordinate of point R is, (x1, y1, z1) and it passes through L2.

${{{x_1} + 1} \over { - 3}} = {{{y_1} - 3} \over 2} = {{{z_1} - 2} \over { - 1}} = t$

$\therefore$   x1 = $-$1 $-$ 3t, y1 = 3 + 2t, z1 = 2 $-$ t

$\overrightarrow {AR} = \left( {3 - 3t} \right)\widehat i + \left( {2t} \right)\widehat j + \left( {1 - t} \right)\widehat k$

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

As $\overrightarrow {AR}$ and $\overrightarrow n$ are perpendicular to each other, So

$\overrightarrow {AR}$ $\cdot$ $\overrightarrow n$ = 0

$\Rightarrow$  (3 $-$ 3t) 1 + (2t)2 + (1 $-$ t) ($-$ 1) = 0

$\Rightarrow$  3 $-$ 3t + 4t $-$ 1 + t = 0

$\Rightarrow$  2 + 2t = 0

$\Rightarrow$  t = $-$ 1

$\therefore$   point R = (2, 1, 3)

$\therefore$  DR of line L is

= (2 $-$ ($-$ 4), $1$ $-$ 3, 3 $-$ 1)

= (6, $-$ 2, 2)

$\therefore$  Equation of line is

${{x + 4} \over 6} = {{y - 3} \over { - 2}} = {{z - 1} \over 2}$

or ${{x + 4} \over 3} = {{y - 3} \over { - 1}} = {{z - 1} \over 1}$