Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

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

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

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

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

2

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

Equation of 1^{st} line is

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

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

Equation of 2^{nd} line is

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

Dr's of 2^{nd} 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

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

Dr's of 1

Equation of 2

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

Dr's of 2

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

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

3

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 :

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

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)

b_{1} $$=$$ $$-$$ 3 and b_{2} $$=$$ 5

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

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

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

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

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)

b

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

4

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

$$\left( {{1 \over 2},4, - 2} \right)$$

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

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

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

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

Number in Brackets after Paper Name Indicates No of Questions

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Trigonometric Functions & Equations *keyboard_arrow_right*

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Complex Numbers *keyboard_arrow_right*

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Vector Algebra and 3D Geometry *keyboard_arrow_right*

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Limits, Continuity and Differentiability *keyboard_arrow_right*

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*