Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

Transportation Engineering

Irrigation

Engineering Mathematics

Construction Material and Management

Fluid Mechanics and Hydraulic Machines

Hydrology

Environmental Engineering

Engineering Mechanics

Structural Analysis

Reinforced Cement Concrete

Steel Structures

Geomatics Engineering Or Surveying

General Aptitude

1

In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$$\widehat i$$ + $$\widehat j$$ $$-$$ $$\widehat k$$, $$-$$$$\widehat i$$ + 3$$\widehat j$$ + p$$\widehat k$$ and 5$$\widehat i$$ + q$$\widehat j$$ $$-$$ 4$$\widehat k$$, then the point (p, q) lies
on a line :

A

parallel to x-axis.

B

parallel to y-axis.

C

making an acute angle with the positive direction of x-axis.

D

making an obtuse angle with the positive direction of x-axis.

Given,

$$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$$

$$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$$

$$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$$

$$ \therefore $$ $$\overrightarrow {AB} = - 4\widehat i + 2\widehat j + \left( {p + 1} \right)\widehat k$$

$$\overrightarrow {AC} = 2\widehat i + \left( {q - 1} \right)\widehat j - 3\widehat k$$

$$\Delta $$ABC is a right angle triangle.

Here $$\overrightarrow {AB} $$ perpendicular to $$\overrightarrow {AC} $$

$$ \therefore $$ $$\overrightarrow {AB} $$ . $$\overrightarrow {AC} $$ = 0

$$ \Rightarrow $$ $$-$$ 8 + 2(q $$-$$ 1) $$-$$ 3(p + 1) = 0

$$ \Rightarrow $$ 3p $$-$$ 2q + 13 = 0

$$ \therefore $$ (p, q) lies on the line

3x $$-$$ 2y + 13 = 0

And slope of the line = $${3 \over 2}$$

$$ \therefore $$ line makes an angle less than 90^{o} or acute angle with the positive direction of x-axis.

$$\overrightarrow A = 3\widehat i + \widehat j - \widehat k$$

$$\overrightarrow B = - \widehat i + 3\widehat j - p\widehat k$$

$$\overrightarrow C = 5\widehat i + 9\widehat j - 4\widehat k$$

$$ \therefore $$ $$\overrightarrow {AB} = - 4\widehat i + 2\widehat j + \left( {p + 1} \right)\widehat k$$

$$\overrightarrow {AC} = 2\widehat i + \left( {q - 1} \right)\widehat j - 3\widehat k$$

$$\Delta $$ABC is a right angle triangle.

Here $$\overrightarrow {AB} $$ perpendicular to $$\overrightarrow {AC} $$

$$ \therefore $$ $$\overrightarrow {AB} $$ . $$\overrightarrow {AC} $$ = 0

$$ \Rightarrow $$ $$-$$ 8 + 2(q $$-$$ 1) $$-$$ 3(p + 1) = 0

$$ \Rightarrow $$ 3p $$-$$ 2q + 13 = 0

$$ \therefore $$ (p, q) lies on the line

3x $$-$$ 2y + 13 = 0

And slope of the line = $${3 \over 2}$$

$$ \therefore $$ line makes an angle less than 90

2

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$$\sqrt 2 $$

3

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

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

4

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.

Number in Brackets after Paper Name Indicates No of Questions

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

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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*

Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *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*