Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Let $$\sqrt 3 \widehat i + \widehat j,$$ $$\widehat i + \sqrt 3 \widehat j$$ and $$\beta \widehat i + \left( {1 - \beta } \right)\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is $${3 \over {\sqrt 2 }}$$, then the sum of all possible values of $$\beta $$ is :

A

4

B

1

C

2

D

3

Angle bisector is x $$-$$ y = 0

$$ \Rightarrow $$ $${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$$

$$ \Rightarrow $$ $$\left| {2\beta - 1} \right| = 3$$

$$ \Rightarrow $$ $$\beta $$ = 2 or $$-$$ 1

$$ \Rightarrow $$ $${{\left| {\beta - \left( {1 - \beta } \right)} \right|} \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$$

$$ \Rightarrow $$ $$\left| {2\beta - 1} \right| = 3$$

$$ \Rightarrow $$ $$\beta $$ = 2 or $$-$$ 1

2

MCQ (Single Correct Answer)

If the point (2, $$\alpha $$, $$\beta $$) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2$$\alpha $$ – 3$$\beta $$ is equal to

A

12

B

7

C

17

D

5

Normal vector of plane

$$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & { - 5} & 0 \cr 4 & { - 4} & 5 \cr } } \right|$$

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

equation of plane is

5(x $$-$$ 7) + 2y $$-$$ 3(z $$-$$ 6) = 0

5x + 2y $$-$$ 3z = 17

5 $$ \times $$ 2 + 2$$\alpha $$ $$-$$ 3$$\beta $$ = 17

$$ \therefore $$ 2$$\alpha $$ $$-$$ 3$$\beta $$ = 17 $$-$$ 10 = 7

$$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & { - 5} & 0 \cr 4 & { - 4} & 5 \cr } } \right|$$

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

equation of plane is

5(x $$-$$ 7) + 2y $$-$$ 3(z $$-$$ 6) = 0

5x + 2y $$-$$ 3z = 17

5 $$ \times $$ 2 + 2$$\alpha $$ $$-$$ 3$$\beta $$ = 17

$$ \therefore $$ 2$$\alpha $$ $$-$$ 3$$\beta $$ = 17 $$-$$ 10 = 7

3

MCQ (Single Correct Answer)

Two lines $${{x - 3} \over 1} = {{y + 1} \over 3} = {{z - 6} \over { - 1}}$$ and $${{x + 5} \over 7} = {{y - 2} \over { - 6}} = {{z - 3} \over 4}$$ intersect at the point R. The reflection of R in the xy-plane has coordinates :

A

(2, 4, 7)

B

(2, $$-$$ 4, $$-$$7)

C

(2, $$-$$ 4, 7)

D

($$-$$ 2, 4, 7)

Point on L_{1} ($$\lambda $$ + 3, 3$$\lambda $$ $$-$$ 1, $$-$$$$\lambda $$ + 6)

Point on L_{2} (7$$\mu $$ $$-$$ 5, $$-$$6$$\mu $$ + 2, 4$$\mu $$ + 3

$$ \Rightarrow $$ $$\lambda $$ + 3 = 7$$\mu $$ $$-$$ 5 . . . . (i)

3$$\lambda $$ $$-$$ 1 = $$-$$6$$\mu $$ + 2 . . . .(ii)

$$ \Rightarrow $$ $$\lambda $$ = $$-$$1, $$\mu $$ = 1

point R(2, $$-$$ 4, 7)

Reflection is (2, $$-$$4, $$-$$ 7)

Point on L

$$ \Rightarrow $$ $$\lambda $$ + 3 = 7$$\mu $$ $$-$$ 5 . . . . (i)

3$$\lambda $$ $$-$$ 1 = $$-$$6$$\mu $$ + 2 . . . .(ii)

$$ \Rightarrow $$ $$\lambda $$ = $$-$$1, $$\mu $$ = 1

point R(2, $$-$$ 4, 7)

Reflection is (2, $$-$$4, $$-$$ 7)

4

MCQ (Single Correct Answer)

The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle $${\pi \over 4}$$ with the plane y $$-$$ z + 5 = 0 are :

A

2, $$-$$1, 1

B

$$2\sqrt 3 ,1, - 1$$

C

$$\sqrt 2 ,1, - 1$$

D

$$\sqrt 2 , - \sqrt 2 $$

Let the equation of plane be

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

It passes through (0, 0, 1) then

b + c = 0 . . . . (1)

Now cos $${\pi \over 4}$$ = $${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {c^2}} }}$$

$$ \Rightarrow $$ a^{2} $$=$$ $$-$$ 2bc and b $$=$$ $$-$$ c

we get a^{2} $$=$$ 2c^{2}

$$ \Rightarrow $$ a $$=$$ $$ \pm $$ $$\sqrt 2 $$ c

$$ \Rightarrow $$ direction ratio (a, b, c) = ($$\sqrt 2 $$, $$-$$1, 1) or ($$\sqrt 2 $$, 1, $$-$$ 1)

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

It passes through (0, 0, 1) then

b + c = 0 . . . . (1)

Now cos $${\pi \over 4}$$ = $${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {c^2}} }}$$

$$ \Rightarrow $$ a

we get a

$$ \Rightarrow $$ a $$=$$ $$ \pm $$ $$\sqrt 2 $$ c

$$ \Rightarrow $$ direction ratio (a, b, c) = ($$\sqrt 2 $$, $$-$$1, 1) or ($$\sqrt 2 $$, 1, $$-$$ 1)

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Complex Numbers

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