1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point :
A
(–3, 0, -1)
B
(3, 2, 1)
C
(3, 3, -1)
D
(–3, 1, 1)

Explanation

The equation of plane

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

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

As plane is parallel to y axis so the normal vector of plane and dot product of $$\widehat j$$ is zero.

$$ \therefore $$  1 + 3$$\lambda $$ = 0

$$ \Rightarrow $$  $$\lambda $$ = $$-$$ $${1 \over 3}$$

$$ \therefore $$  So the equation of the plane is

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

$$ \Rightarrow $$  x ($${1 \over 3}$$) + z($${4 \over 3}$$) $$-$$ $${7 \over 3}$$ = 0

$$ \Rightarrow $$  x + 4z $$-$$ 7 = 0

By checking each options you can see only point (3, 2, 1) lies on the plane.
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

The equation of the plane containing the straight line $${x \over 2} = {y \over 3} = {z \over 4}$$ and perpendicular to the plane containing the straight lines $${x \over 3} = {y \over 4} = {z \over 2}$$ and $${x \over 4} = {y \over 2} = {z \over 3}$$ is :
A
x $$-$$ 2y + z = 0
B
3x + 2y $$-$$ 3z = 0
C
x + 2y $$-$$ 2z = 0
D
5x + 2y $$-$$ 4z = 0

Explanation

Vector $$ \bot $$ to given plane = $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 4 & 2 \cr 4 & 2 & 3 \cr } } \right|$$

= $$\widehat i\left( {12 - 4} \right) - \widehat j\left( {9 - 8} \right) + \widehat k\left( {6 - 16} \right)$$

= $$8\widehat i - \widehat j - 10\widehat k\,$$      . . . . (1)

Vector parallel to given line

= $$2\widehat i + 3\widehat j + 4\widehat k\,\,\,\,$$      . . . (2)

Vector $$ \bot \,\,\,$$ to both (1) & (2) vector

= $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 8 & { - 1} & { - 10} \cr 2 & 3 & 4 \cr } } \right|$$

= $$\widehat i\left( { - 4 + 30} \right) - \widehat j\left( {32 + 20} \right) + \widehat k\left( {24 + 2} \right)$$

= $$26\widehat i - 52\widehat j + 26\widehat k$$

Dr's of normal of required plane is

(26, $$-$$52, 26) $$ \Rightarrow $$ (1, $$-$$2, 1)

Equation of plane whose Dr's of Normal is (1, $$-$$2, 1) and passes through origin

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

x $$-$$ 2y + z = 0
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

If the lines x = ay + b, z = cy + d and x = a'z + b', y = c'z + d' are perpendicular, then :
A
ab'  +  bc'  +  1  =  0
B
cc'  +  a   +  a'  =  0
C
bb'  +  cc'  +  1  =  0
D
aa'  +  c  +  c'  =  0

Explanation

Equation of 1st line is

$${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$$

Dr's of 1st line = ($$a$$, 1 , c)

Equation of 2nd line is

$${{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}$$

Dr's of 2nd line = ($$a'$$, c' , 1)

Lines are perpendicular, so the dot product of the Dr's of two lines are zero.

$$ \therefore $$ $$aa$$' + c + c' = 0
4
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

$$ \Rightarrow $$  $$\left| {\overrightarrow b } \right| = \sqrt {9 + 25 + 2} = 6$$

Questions Asked from Vector Algebra and 3D Geometry

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