1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

Let  $$\overrightarrow a = \widehat i + \widehat j + \sqrt 2 \widehat k,$$   $$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + \sqrt 2 \widehat k$$,    $$\overrightarrow c = 5\widehat i + \widehat j + \sqrt 2 \widehat k$$   be three vectors such that the projection vector of $$\overrightarrow b $$ on $$\overrightarrow a $$ is $$\overrightarrow a $$.
If   $$\overrightarrow a + \overrightarrow b $$   is perpendicular to $$\overrightarrow c $$ , then $$\left| {\overrightarrow b } \right|$$ is equal to :
A
$$\sqrt {32} $$
B
6
C
$$\sqrt {22} $$
D
4

Explanation

Projection of $$\overrightarrow b $$ on $$\overrightarrow a $$ is $$\overrightarrow a $$

$$ \therefore $$   $${{\overrightarrow b \cdot \overrightarrow a } \over {\left| {\overrightarrow a } \right|}} = \left| {\overrightarrow a } \right|$$

$$ \Rightarrow $$  $${{{b_1} + {b_2} + 2} \over 2} = 2$$

$$ \Rightarrow $$  $${b_1} + {b_2} = 2\,\,\,\,\,\,\,\,\,\,\,....(1)$$

and $$\overrightarrow a $$ + $$\overrightarrow b $$ is perpendicular to $$\overrightarrow c $$

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

$$ \Rightarrow $$  $$5\left( {{b_1} + 1} \right) + \left( {{b_2} + 1} \right) + \sqrt 2 \left( {2\sqrt 2 } \right) = 0$$

$$ \Rightarrow $$  $$5{b_1} + {b_2} + 10 = 0\,\,\,\,\,\,\,\,\,\,......(2)$$

solving (1) & (2)

b1 $$=$$ $$-$$ 3 and b2 $$=$$ 5

$$ \Rightarrow $$  $$\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$$
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

Let $$\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k,\,\,$$   $$\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k,$$  and  $$\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$$  be three vectors such that $$\overrightarrow b = 2\overrightarrow a $$ and $$\overrightarrow a $$ is perpendicular to $$\overrightarrow c $$. Then a possible value of $$\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right)$$ is -
A
(1, 5, 1)
B
(1, 3, 1)
C
$$\left( { - {1 \over 2},4,0} \right)$$
D
$$\left( {{1 \over 2},4, - 2} \right)$$

Explanation

Given $$\overrightarrow b = 2\overrightarrow a $$

$$ \therefore $$ $$4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$$

$$ \Rightarrow 3 - {\lambda _2} = 2{\lambda _1} \Rightarrow 2{\lambda _1} + {\lambda _2} = 3\,\,...(1)$$

Given $$\overrightarrow a $$ is perpendicular to $$\overrightarrow c $$

$$ \therefore $$ $$\overrightarrow a .\overrightarrow c = 0$$

$$ \Rightarrow 6 + 6{\lambda _1} + 3\left( {{\lambda _3} - 1} \right) = 0$$

$$ \Rightarrow 2{\lambda _1} + {\lambda _3} = - 1\,\,\,\,\,\,\,\,\,\,...(2)$$

Now $$\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right) = \left( {{\lambda _1},3 - 2{\lambda _1}, - 1 - 2{\lambda _1}} \right)$$

By checking each option you can see,

when $${\lambda _1}$$ = $$ - {1 \over 2}$$

then $${\lambda _2}$$ = $$3 - 2{\lambda _1}$$ = 3 + 1 = 4

and $${\lambda _3}$$ = $$-1 - 2{\lambda _1}$$ = - 1 + 1 = 0
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

The plane passing through the point (4, –1, 2) and parallel to the lines  $${{x + 2} \over 3} = {{y - 2} \over { - 1}} = {{z + 1} \over 2}$$  and  $${{x - 2} \over 1} = {{y - 3} \over 2} = {{z - 4} \over 3}$$ also passes through the point -
A
(1, 1, $$-$$ 1)
B
(1, 1, 1)
C
($$-$$ 1, $$-$$ 1, $$-$$1)
D
($$-$$ 1, $$-$$ 1, 1)

Explanation

Let $$\overrightarrow n $$ be the normal vector to the plane passing through (4, $$-$$1, 2) and parallel to the lines L1 & L2

then $$\overrightarrow n $$ = $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 2 \cr 1 & 2 & 3 \cr } } \right|$$

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

$$ \therefore $$  Equation of plane is

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

$$ \therefore $$  x + y $$-$$ z $$-$$ 1 = 0

Now check options
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

Let A be a point on the line $$\overrightarrow r = \left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$ and B(3, 2, 6) be a point in the space. Then the value of $$\mu $$ for which the vector $$\overrightarrow {AB} $$  is parallel to the plane x $$-$$ 4y + 3z = 1 is -
A
$${1 \over 8}$$
B
$${1 \over 2}$$
C
$${1 \over 4}$$
D
$$-$$ $${1 \over 4}$$

Explanation

Let point A is

$$\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$

and point B is (3, 2, 6)

then $$\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\widehat k$$

which is parallel to the the plane x $$-$$ 4y + 3z = 1

$$ \therefore $$  2 + 3$$\mu $$ $$-$$ 12 + 4$$\mu $$ + 12 $$-$$ 15$$\mu $$ = 0

8$$\mu $$ = 2

$$\mu $$ = $${1 \over 4}$$

Questions Asked from Vector Algebra and 3D Geometry

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