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 distance of the point (1, − 2, 4) from the plane passing through the point
(1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x − 2y + z + 12 = 0, is :

A

$$2\sqrt 2 $$

B

2

C

$$\sqrt 2 $$

D

$${1 \over {\sqrt 2 }}$$

Equation of plane passing through point (1, 2, 2) is,

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

This plane is perpendicular to the plane

x $$-$$ y + 2z = 3 and 2x $$-$$ 2y + z + 12 = 0

When two planes,

a_{1}x + b_{1}y + c_{1}z + d_{1} = 0

and a_{2}x + b_{2}y + c_{2}z + d_{2} = 0

are perpendicular to each other then

a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

So, we can say,

a $$ \times $$ 1 + b $$ \times $$ ($$-$$ 1) + c $$ \times $$ 2 = 0

$$ \Rightarrow $$ a $$-$$ b + 2c = 0 . . . (2)

and, a $$ \times $$ 2 + b($$-$$2) + c(1) = 0

$$ \Rightarrow $$ 2a $$-$$ 2b + c = 0 . . .(3)

Solving (2) and (3)

$${a \over { - 1 + 4}}$$ = $${b \over {4 - 1}}$$ = $${c \over { - 2 + 2}}$$ = $$\lambda $$

$$ \Rightarrow $$ a = 3$$\lambda $$, b = 3$$\lambda $$, c = 0

Putting the values of a, b, c in equation (1) we get,

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

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

$$ \Rightarrow $$ 3x + 3y $$-$$ 9 = 0

$$ \Rightarrow $$ x + y $$-$$ 3 = 0

$$ \therefore $$ Distance of point (1, $$-$$2, 4) from plane x + y $$-$$ 3 = 0 is,

= $$\left| {{{1 - 2 - 3} \over {\sqrt {1 + 1} }}} \right|$$ = $${4 \over {\sqrt 2 }}$$ = 2$$\sqrt 2 $$

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

This plane is perpendicular to the plane

x $$-$$ y + 2z = 3 and 2x $$-$$ 2y + z + 12 = 0

When two planes,

a

and a

are perpendicular to each other then

a

So, we can say,

a $$ \times $$ 1 + b $$ \times $$ ($$-$$ 1) + c $$ \times $$ 2 = 0

$$ \Rightarrow $$ a $$-$$ b + 2c = 0 . . . (2)

and, a $$ \times $$ 2 + b($$-$$2) + c(1) = 0

$$ \Rightarrow $$ 2a $$-$$ 2b + c = 0 . . .(3)

Solving (2) and (3)

$${a \over { - 1 + 4}}$$ = $${b \over {4 - 1}}$$ = $${c \over { - 2 + 2}}$$ = $$\lambda $$

$$ \Rightarrow $$ a = 3$$\lambda $$, b = 3$$\lambda $$, c = 0

Putting the values of a, b, c in equation (1) we get,

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

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

$$ \Rightarrow $$ 3x + 3y $$-$$ 9 = 0

$$ \Rightarrow $$ x + y $$-$$ 3 = 0

$$ \therefore $$ Distance of point (1, $$-$$2, 4) from plane x + y $$-$$ 3 = 0 is,

= $$\left| {{{1 - 2 - 3} \over {\sqrt {1 + 1} }}} \right|$$ = $${4 \over {\sqrt 2 }}$$ = 2$$\sqrt 2 $$

2

ABC is a triangle in a plane with vertices

A(2, 3, 5), B(−1, 3, 2) and C($$\lambda $$, 5, $$\mu $$).

If the median through A is equally inclined to the coordinate axes, then the value of ($$\lambda $$^{3} + $$\mu $$^{3} + 5) is :

A(2, 3, 5), B(−1, 3, 2) and C($$\lambda $$, 5, $$\mu $$).

If the median through A is equally inclined to the coordinate axes, then the value of ($$\lambda $$

A

1130

B

1348

C

676

D

1077

DR's of AD are

$${{\lambda - 1} \over 2} - 2,{{5 + 3} \over 2} - 3,{{\mu + 2} \over 2} - 5$$

i.e. $${{\lambda - 5} \over 2},\,\,1,\,\,{{\mu - 8} \over 2}$$

As medium is making equal angles with coordinate axes,

$$ \therefore $$ $${{\lambda - 5} \over 2} = 1 = {{\mu - 8} \over 2}$$

$$ \Rightarrow $$ $$\lambda $$ = 7, $$\mu $$ = 10

$$ \therefore $$ $$\lambda $$

3

The number of distinct real values of $$\lambda $$ for which the lines

$${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over {{\lambda ^2}}}$$ and $${{x - 3} \over 1} = {{y - 2} \over {{\lambda ^2}}} = {{z - 1} \over 2}$$ are coplanar is :

$${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over {{\lambda ^2}}}$$ and $${{x - 3} \over 1} = {{y - 2} \over {{\lambda ^2}}} = {{z - 1} \over 2}$$ are coplanar is :

A

4

B

1

C

2

D

3

As planes are coplanar, so

$$\left| {\matrix{ {3 - 1} & {2 - 2} & {1 - \left( { - 3} \right)} \cr 1 & 2 & {{\lambda ^2}} \cr 1 & {{\lambda ^2}} & 2 \cr } } \right| $$ = 0

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

$$ \Rightarrow $$ 2(4 $$-$$ $$\lambda $$^{4}) + 4($$\lambda $$^{2} $$-$$ 2) = 0

$$ \Rightarrow $$ 4 $$-$$ $$\lambda $$^{4} + 2$$\lambda $$^{2} $$-$$ 4 = 0

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

$$ \Rightarrow $$ $$\lambda $$ = 0, $$\sqrt 2 , - \sqrt 2 $$

$$ \therefore $$ 3 distinct real values are possible.

$$\left| {\matrix{ {3 - 1} & {2 - 2} & {1 - \left( { - 3} \right)} \cr 1 & 2 & {{\lambda ^2}} \cr 1 & {{\lambda ^2}} & 2 \cr } } \right| $$ = 0

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

$$ \Rightarrow $$ 2(4 $$-$$ $$\lambda $$

$$ \Rightarrow $$ 4 $$-$$ $$\lambda $$

$$ \Rightarrow $$ $$\lambda $$

$$ \Rightarrow $$ $$\lambda $$ = 0, $$\sqrt 2 , - \sqrt 2 $$

$$ \therefore $$ 3 distinct real values are possible.

4

Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$ respectively, then the position vector of the orthocentre of this triangle, is :

A

$${\overrightarrow a + \overrightarrow b + \overrightarrow c }$$

B

$$ - \left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$$

C

$$\overrightarrow 0 $$

D

$$\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$$

Given,

Position vector of circumcentre, $$\overrightarrow C = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$

We know, position vector of centroid, $$\overrightarrow G = {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 3}$$

Now, let $$\overrightarrow R $$ be the orthocentre of the triangle.

We know, $$\overrightarrow G $$ $$ = {{2\overrightarrow C + \overrightarrow R } \over 3}$$

$$ \Rightarrow $$ 3$$\overrightarrow G $$ $$ = 2\overrightarrow C + \overrightarrow R $$

$$ \Rightarrow $$ $$\overrightarrow R = 3\overrightarrow G - 2\overrightarrow C $$

= $$\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) - 2\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}} \right)$$

= $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}$$

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (9) *keyboard_arrow_right*

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*