Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

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Construction Material and Management

Reinforced Cement Concrete

Steel Structures

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Geomatics Engineering Or Surveying

Environmental Engineering

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Engineering Mathematics

General Aptitude

1

The shortest distance between the lines $${x \over 2} = {y \over 2} = {z \over 1}$$ and

$${{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}$$ lies in the interval :

$${{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}$$ lies in the interval :

A

[0, 1)

B

[1, 2)

C

(2, 3]

D

(3, 4]

Shortest distance between the lines

$${{x - {x_1}} \over {{a_1}}} = {{y - {y_1}} \over {{b_1}}} = {{z - {z_1}} \over {{c_1}}}$$

and $${{x - {x_2}} \over {{a_2}}} = {{y - {y_2}} \over {{b_2}}} = {{z - {z_2}} \over {{c_2}}}$$ is

$$\left| {{{\left| {\matrix{ {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{a_1}} & {{b_1}} & {{c_1}} \cr {{a_2}} & {{b_2}} & {{c_2}} \cr } } \right|} \over {\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}} \right|$$

$$ \therefore $$ Shortest distance between two given lines are,

$$\left| {{{\left| {\matrix{ { - 2} & 4 & 5 \cr 2 & 2 & 1 \cr { - 1} & 8 & 4 \cr } } \right|} \over {\sqrt {{{\left( {8 - 8} \right)}^2} + {{\left( { - 1 - 8} \right)}^2} + {{\left( {16 + 2} \right)}^2}} }}} \right|$$

= $$\left| {{{ - 36 + 90} \over {\sqrt {405} }}} \right|$$

= $${{54} \over {20.1}}$$

= 2.68

$${{x - {x_1}} \over {{a_1}}} = {{y - {y_1}} \over {{b_1}}} = {{z - {z_1}} \over {{c_1}}}$$

and $${{x - {x_2}} \over {{a_2}}} = {{y - {y_2}} \over {{b_2}}} = {{z - {z_2}} \over {{c_2}}}$$ is

$$\left| {{{\left| {\matrix{ {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{a_1}} & {{b_1}} & {{c_1}} \cr {{a_2}} & {{b_2}} & {{c_2}} \cr } } \right|} \over {\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}} \right|$$

$$ \therefore $$ Shortest distance between two given lines are,

$$\left| {{{\left| {\matrix{ { - 2} & 4 & 5 \cr 2 & 2 & 1 \cr { - 1} & 8 & 4 \cr } } \right|} \over {\sqrt {{{\left( {8 - 8} \right)}^2} + {{\left( { - 1 - 8} \right)}^2} + {{\left( {16 + 2} \right)}^2}} }}} \right|$$

= $$\left| {{{ - 36 + 90} \over {\sqrt {405} }}} \right|$$

= $${{54} \over {20.1}}$$

= 2.68

2

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

3

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$$\sqrt 2 $$

4

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

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

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