Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

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)

2

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)

3

MCQ (Single Correct Answer)

Let $$\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$$ $$\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$$ and $$\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$$ be coplanar vectors. Then the non-zero vector $$\overrightarrow a \times \overrightarrow c $$ is :

A

$$ - 10\widehat i - 5\widehat j$$

B

$$ - 10\widehat i + 5\widehat j$$

C

$$ - 14\widehat i + 5\widehat j$$

D

$$ - 14\widehat i - 5\widehat j$$

$$\left[ {\overrightarrow a \,\,\overrightarrow b \,\,\overrightarrow c } \right] = 0$$

$$ \Rightarrow \left| {\matrix{ 1 & 2 & 4 \cr 1 & \lambda & 4 \cr 2 & 4 & {{\lambda ^2} - 1} \cr } } \right| = 0$$

$$ \Rightarrow {\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0$$

$$ \Rightarrow {\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0$$

$$ \Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0$$

$$ \Rightarrow \lambda = 2,3, - 3$$

So, $$\lambda $$ = 2 (as $$\overrightarrow a $$ is parallel to $$\overrightarrow c $$ for $$\lambda $$ = $$ \pm $$3)

Hence $$\overrightarrow a \times \overrightarrow c = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 4 \cr 2 & 4 & 3 \cr } } \right|$$

$$ = - 10\widehat i + 5\widehat j$$

$$ \Rightarrow \left| {\matrix{ 1 & 2 & 4 \cr 1 & \lambda & 4 \cr 2 & 4 & {{\lambda ^2} - 1} \cr } } \right| = 0$$

$$ \Rightarrow {\lambda ^3} - 2{\lambda ^2} - 9\lambda + 18 = 0$$

$$ \Rightarrow {\lambda ^2}\left( {\lambda - 2} \right) - 9\left( {\lambda - 2} \right) = 0$$

$$ \Rightarrow \left( {\lambda - 3} \right)\left( {\lambda + 3} \right)\left( {\lambda - 2} \right) = 0$$

$$ \Rightarrow \lambda = 2,3, - 3$$

So, $$\lambda $$ = 2 (as $$\overrightarrow a $$ is parallel to $$\overrightarrow c $$ for $$\lambda $$ = $$ \pm $$3)

Hence $$\overrightarrow a \times \overrightarrow c = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 4 \cr 2 & 4 & 3 \cr } } \right|$$

$$ = - 10\widehat i + 5\widehat j$$

4

MCQ (Single Correct Answer)

The plane containing the line $${{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z - 1} \over 3}$$ and also containing its projection on the plane 2x + 3y $$-$$ z = 5, contains which one of the following points ?

A

($$-$$ 2, 2, 2)

B

(2, 2, 0)

C

(2, 0, $$-$$ 2)

D

(0, $$-$$ 2, 2)

The normal vector of required plane

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

$$ = - 8\widehat i + 8\widehat j + 8\widehat k$$

So, direction ratio of normal is $$\left( { - 1,1,1} \right)$$

So required plane is

$$ - \left( {x - 3} \right) + \left( {y + 2} \right) + \left( {z - 1} \right) = 0$$

$$ \Rightarrow - x + y + z + 4 = 0$$

Which is satisfied by $$\left( {2,0, - 2} \right)$$

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

$$ = - 8\widehat i + 8\widehat j + 8\widehat k$$

So, direction ratio of normal is $$\left( { - 1,1,1} \right)$$

So required plane is

$$ - \left( {x - 3} \right) + \left( {y + 2} \right) + \left( {z - 1} \right) = 0$$

$$ \Rightarrow - x + y + z + 4 = 0$$

Which is satisfied by $$\left( {2,0, - 2} \right)$$

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Complex Numbers

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