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1

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
Two lines $${{x - 3} \over 1} = {{y + 1} \over 3} = {{z - 6} \over { - 1}}$$ and $${{x + 5} \over 7} = {{y - 2} \over { - 6}} = {{z - 3} \over 4}$$ intersect at the point R. The reflection of R in the xy-plane has coordinates :
A
(2, 4, 7)
B
(2, $$-$$ 4, $$-$$7)
C
(2, $$-$$ 4, 7)
D
($$-$$ 2, 4, 7)

Explanation

Point on L1 ($$\lambda $$ + 3, 3$$\lambda $$ $$-$$ 1, $$-$$$$\lambda $$ + 6)

Point on L2 (7$$\mu $$ $$-$$ 5, $$-$$6$$\mu $$ + 2, 4$$\mu $$ + 3

$$ \Rightarrow $$  $$\lambda $$ + 3 = 7$$\mu $$ $$-$$ 5      . . . . (i)

3$$\lambda $$ $$-$$ 1 = $$-$$6$$\mu $$ + 2       . . . .(ii)

$$ \Rightarrow $$  $$\lambda $$ = $$-$$1, $$\mu $$ = 1

point R(2, $$-$$ 4, 7)

Reflection is (2, $$-$$4, $$-$$ 7)
2

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)
The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y $$-$$ z + 5 = 0 are :
A
2, $$-$$1, 1
B
$$2\sqrt 3 ,1, - 1$$
C
$$\sqrt 2 ,1, - 1$$
D
$$\sqrt 2 , - \sqrt 2 $$

Explanation

Let the equation of plane be

a(x $$-$$ 0) + b(y + 1) + c(z $$-$$ 0) = 0

It passes through (0, 0, 1) then

b + c = 0     . . . . (1)

Now cos $${\pi \over 4}$$ = $${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {c^2}} }}$$

$$ \Rightarrow $$  a2 $$=$$ $$-$$ 2bc and b $$=$$ $$-$$ c

we get a2 $$=$$ 2c2

$$ \Rightarrow $$  a $$=$$ $$ \pm $$ $$\sqrt 2 $$ c

$$ \Rightarrow $$  direction ratio (a, b, c) = ($$\sqrt 2 $$, $$-$$1, 1) or ($$\sqrt 2 $$, 1, $$-$$ 1)
3

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)
Let  $$\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$$ $$\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$$ and $$\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$$ be coplanar vectors. Then the non-zero vector $$\overrightarrow a \times \overrightarrow c $$ is :
A
$$ - 10\widehat i - 5\widehat j$$
B
$$ - 10\widehat i + 5\widehat j$$
C
$$ - 14\widehat i + 5\widehat j$$
D
$$ - 14\widehat i - 5\widehat j$$

Explanation

$$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0$$

$$ \Rightarrow \left| {\matrix{ 1 & 2 & 4 \cr 1 & \lambda & 4 \cr 2 & 4 & {{\lambda ^2} - 1} \cr } } \right| = 0$$

$$ \Rightarrow {\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0$$

$$ \Rightarrow {\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0$$

$$ \Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0$$

$$ \Rightarrow \lambda = 2,3, - 3$$

So,   $$\lambda $$ = 2 (as   $$\overrightarrow a $$ is parallel to $$\overrightarrow c $$ for $$\lambda $$ = $$ \pm $$3)

Hence   $$\overrightarrow a \times \overrightarrow c = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 4 \cr 2 & 4 & 3 \cr } } \right|$$

$$ = - 10\widehat i + 5\widehat j$$
4

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)
The plane containing the line $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$$ and also containing its projection on the plane 2x + 3y $$-$$ z = 5, contains which one of the following points ?
A
($$-$$ 2, 2, 2)
B
(2, 2, 0)
C
(2, 0, $$-$$ 2)
D
(0, $$-$$ 2, 2)

Explanation

The normal vector of required plane

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

$$ = - 8\widehat i + 8\widehat j + 8\widehat k$$

So, direction ratio of normal is $$\left( { - 1,1,1} \right)$$

So required plane is

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

$$ \Rightarrow - x + y + z + 4 = 0$$

Which is satisfied by $$\left( {2,0, - 2} \right)$$

Questions Asked from Vector Algebra and 3D Geometry

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