Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The sum of the distinct real values of $$\mu $$, for which the vectors, $$\mu \widehat i + \widehat j + \widehat k,$$ $$\widehat i + \mu \widehat j + \widehat k,$$ $$\widehat i + \widehat j + \mu \widehat k$$ are co-planar, is :

A

2

B

$$-$$1

C

0

D

1

$$\left| {\matrix{
\mu & 1 & 1 \cr
1 & \mu & 1 \cr
1 & 1 & \mu \cr
} } \right| = 0$$

$$\mu \left( {{\mu ^2} - 1} \right) - 1\left( {\mu - 1} \right) + 1\left( {1 - \mu } \right) = 0$$

$${\mu ^3} - \mu - \mu + 1 + 1\mu = 0$$

$${\mu ^3} - 3\mu + 2 = 0$$

$${\mu ^3} - 1 - 3\left( {\mu - 1} \right) = 0$$

$$u = 1,\,\,{\mu ^2} + \mu - 2 = 0$$

$$\mu = 1,\,\,\mu = - 2$$

sum of distinct solutions $$=$$ $$-$$ 1

$$\mu \left( {{\mu ^2} - 1} \right) - 1\left( {\mu - 1} \right) + 1\left( {1 - \mu } \right) = 0$$

$${\mu ^3} - \mu - \mu + 1 + 1\mu = 0$$

$${\mu ^3} - 3\mu + 2 = 0$$

$${\mu ^3} - 1 - 3\left( {\mu - 1} \right) = 0$$

$$u = 1,\,\,{\mu ^2} + \mu - 2 = 0$$

$$\mu = 1,\,\,\mu = - 2$$

sum of distinct solutions $$=$$ $$-$$ 1

2

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

3

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 â€“ 5y = 15, then 2$$\alpha $$ â€“ 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

4

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)

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