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

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

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

2

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

3

The plane passing through the point (4, –1, 2) and parallel to the lines $${{x + 2} \over 3} = {{y - 2} \over { - 1}} = {{z + 1} \over 2}$$ and $${{x - 2} \over 1} = {{y - 3} \over 2} = {{z - 4} \over 3}$$ also passes through the point -

A

(1, 1, $$-$$ 1)

B

(1, 1, 1)

C

($$-$$ 1, $$-$$ 1, $$-$$1)

D

($$-$$ 1, $$-$$ 1, 1)

Let $$\overrightarrow n $$ be the normal vector to the plane passing through (4, $$-$$1, 2) and parallel to the lines L_{1} & L_{2}

then $$\overrightarrow n $$ = $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 2 \cr 1 & 2 & 3 \cr } } \right|$$

$$ \therefore $$ $$\overrightarrow n $$ = $$ - 7\widehat i - 7\widehat j + 7\widehat k$$

$$ \therefore $$ Equation of plane is

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

$$ \therefore $$ x + y $$-$$ z $$-$$ 1 = 0

Now check options

then $$\overrightarrow n $$ = $$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 2 \cr 1 & 2 & 3 \cr } } \right|$$

$$ \therefore $$ $$\overrightarrow n $$ = $$ - 7\widehat i - 7\widehat j + 7\widehat k$$

$$ \therefore $$ Equation of plane is

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

$$ \therefore $$ x + y $$-$$ z $$-$$ 1 = 0

Now check options

4

Let A be a point on the line $$\overrightarrow r = \left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$ and B(3, 2, 6) be a point in the space. Then the value of $$\mu $$ for which the vector $$\overrightarrow {AB} $$ is parallel to the plane x $$-$$ 4y + 3z = 1 is -

A

$${1 \over 8}$$

B

$${1 \over 2}$$

C

$${1 \over 4}$$

D

$$-$$ $${1 \over 4}$$

Let point A is

$$\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$

and point B is (3, 2, 6)

then $$\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\widehat k$$

which is parallel to the the plane x $$-$$ 4y + 3z = 1

$$ \therefore $$ 2 + 3$$\mu $$ $$-$$ 12 + 4$$\mu $$ + 12 $$-$$ 15$$\mu $$ = 0

8$$\mu $$ = 2

$$\mu $$ = $${1 \over 4}$$

$$\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$$

and point B is (3, 2, 6)

then $$\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\widehat k$$

which is parallel to the the plane x $$-$$ 4y + 3z = 1

$$ \therefore $$ 2 + 3$$\mu $$ $$-$$ 12 + 4$$\mu $$ + 12 $$-$$ 15$$\mu $$ = 0

8$$\mu $$ = 2

$$\mu $$ = $${1 \over 4}$$

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