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1

JEE Main 2021 (Online) 27th August Evening Shift

MCQ (Single Correct Answer)
The equation of the plane passing through the line of intersection of the planes $$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$$ and $$\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0$$ and parallel to the x-axis is :
A
$$\overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0$$
B
$$\overrightarrow r .\left( {\widehat i + 3\widehat k} \right) + 6 = 0$$
C
$$\overrightarrow r .\left( {\widehat i - 3\widehat k} \right) + 6 = 0$$
D
$$\overrightarrow r .\left( {\widehat j - 3\widehat k} \right) - 6 = 0$$

Explanation

Equation of planes are

$$\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) - 1 = 0 \Rightarrow x + y + z - 1 = 0$$

and $$\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0 \Rightarrow 2x + 3y - z + 4 = 0$$

equation of planes through line of intersection of these planes is :-

$$(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0$$

$$ \Rightarrow (1 + 2\lambda )x + (1 + 3\lambda )y + (1 - \lambda )z - 1 + 4\lambda = 0$$

But this plane is parallel to x-axis whose direction are (1, 0, 0)

$$\therefore$$ $$(1 + 2\lambda )1 + (1 + 3\lambda )0 + (1 - \lambda )0 = 0$$

$$\lambda = - {1 \over 2}$$

$$\therefore$$ Required plane is

$$0x + \left( {1 - {3 \over 2}} \right)y + \left( {1 + {1 \over 2}} \right)z - 1 + 4\left( {{{ - 1} \over 2}} \right) = 0$$

$$ \Rightarrow {{ - y} \over 2} + {3 \over 2}z - 3 = 0$$

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

$$ \Rightarrow \overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0$$ Ans.
2

JEE Main 2021 (Online) 27th August Morning Shift

MCQ (Single Correct Answer)
Equation of a plane at a distance $${2 \over {\sqrt {21} }}$$ from the origin, which contains the line of intersection of the planes x $$-$$ y $$-$$ z $$-$$ 1 = 0 and 2x + y $$-$$ 3z + 4 = 0, is :
A
$$3x - y - 5z + 2 = 0$$
B
$$3x - 4z + 3 = 0$$
C
$$ - x + 2y + 2z - 3 = 0$$
D
$$4x - y - 5z + 2 = 0$$

Explanation

Required equation of plane

$${P_1} + \lambda {P_2} = 0$$

$$(x - y - z - 1) + \lambda (2x + y - 3z + 4) = 0$$

Given that its dist. From origin is $${2 \over {\sqrt {21} }}$$

Thus, $${{|4\lambda - 1|} \over {\sqrt {{{(2\lambda + 1)}^2} + {{(\lambda - 1)}^2} + {{( - 3\lambda - 1)}^2}} }} = {{\sqrt 2 } \over {\sqrt {21} }}$$

$$ \Rightarrow 21{(4\lambda - 1)^2} = 2(14{\lambda ^2} + 8\lambda + 3)$$

$$ \Rightarrow 336{\lambda ^2} - 168\lambda + 21 = 28{\lambda ^2} + 16\lambda + 6$$

$$ \Rightarrow 308{\lambda ^2} - 184\lambda + 15 = 0$$

$$ \Rightarrow 308{\lambda ^2} - 154\lambda - 30\lambda + 15 = 0$$

$$ \Rightarrow (2\lambda - 1)(154\lambda - 15) = 0$$

$$ \Rightarrow \lambda = {1 \over 2}$$ or $${{15} \over {154}}$$

for $$\lambda = {1 \over 2}$$ reqd. plane is $$4x - y - 5z + 2 = 0$$
3

JEE Main 2021 (Online) 26th August Evening Shift

MCQ (Single Correct Answer)
Let P be the plane passing through the point (1, 2, 3) and the line of intersection of the planes $$\overrightarrow r \,.\,\left( {\widehat i + \widehat j + 4\widehat k} \right) = 16$$ and $$\overrightarrow r \,.\,\left( { - \widehat i + \widehat j + \widehat k} \right) = 6$$. Then which of the following points does NOT lie on P?
A
(3, 3, 2)
B
(6, $$-$$6, 2)
C
(4, 2, 2)
D
($$-$$8, 8, 6)

Explanation

$$(x + y + 4z - 16) + \lambda ( - x + y + z - 6) = 0$$

Passes through (1, 2, 3)

$$ - 1 + \lambda ( - 2) \Rightarrow \lambda = - {1 \over 2}$$

$$2(x + y + 4z - 16) - ( - x + y + z - 6) = 0$$

$$3x + y + 7z - 26 = 0$$
4

JEE Main 2021 (Online) 26th August Evening Shift

MCQ (Single Correct Answer)
A hall has a square floor of dimension 10 m $$\times$$ 10 m (see the figure) and vertical walls. If the angle GPH between the diagonals AG and BH is $${\cos ^{ - 1}}{1 \over 5}$$, then the height of the hall (in meters) is :

A
5
B
2$$\sqrt {10} $$
C
5$$\sqrt {3} $$
D
5$$\sqrt {2} $$

Explanation

$$A(\widehat j)\,.\,B(10\widehat i)$$

$$H(h\widehat j + 10\widehat k)$$

$$G(10\widehat i + h\widehat j + 10\widehat k)$$

$$\overrightarrow {AG} = 10\widehat i + h\widehat j + 10\widehat k$$

$$\overrightarrow {BH} = - 10\widehat i + h\widehat j + 10\widehat k$$

$$\cos \theta = {{\overrightarrow {AG} \overrightarrow {BH} } \over {\left| {\overrightarrow {AG} } \right|\left| {\overrightarrow {BH} } \right|}}$$

$${1 \over 5} = {{{h^2}} \over {{h^2} + 200}}$$

$$4{h^2} = 200 \Rightarrow h = 5\sqrt 2 $$

Questions Asked from Vector Algebra and 3D Geometry

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