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Graduate Aptitude Test in Engineering

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General Aptitude

1

An angle between the lines whose direction cosines are gien by the equations,

$$l$$ + 3m + 5n = 0 and 5$$l$$m $$-$$ 2mn + 6n$$l$$ = 0, is :

$$l$$ + 3m + 5n = 0 and 5$$l$$m $$-$$ 2mn + 6n$$l$$ = 0, is :

A

$${\cos ^{ - 1}}\left( {{1 \over 3}} \right)$$

B

$${\cos ^{ - 1}}\left( {{1 \over 4}} \right)$$

C

$${\cos ^{ - 1}}\left( {{1 \over 6}} \right)$$

D

$${\cos ^{ - 1}}\left( {{1 \over 8}} \right)$$

Given

l + 3m + 5n = 0

and 5$$l$$m $$-$$ 2mn + 6n$$l$$ = 0

From eq. (1) we have

$$l$$ = $$-$$ 3m $$-$$ 5n

Put the value of $$l$$ in eq. (2), we get ;

5 ($$-$$3m $$-$$5n) m $$-$$ 2mn + 6n ($$-$$ 3m $$-$$ 5n) = 0

$$ \Rightarrow $$ 15m^{2} + 45mn + 30n^{2} = 0

$$ \Rightarrow $$ m^{2} + 3mn + 2n^{2} = 0

$$ \Rightarrow $$ m^{2} + 2mn + mn + 2n^{2} = 0

$$ \Rightarrow $$ $$\,\,\,$$ (m + n) (m + 2n) = 0

$$ \therefore $$ m = $$-$$ n or m = $$-$$ 2n

For m = $$-n, $$ $$l$$ = $$-$$ 2n

And for m = $$-$$ 2n, $$l$$ = n

$$ \therefore $$ ($$l$$, m, n) = ($$-$$2n, $$-$$n, n) Or ($$l$$, m, n) = (n, $$-$$ 2n, n)

$$ \Rightarrow $$ ($$l$$, m, n) = ($$-$$2, $$-$$1, 1) Or ($$l$$, m, n) = (1, $$-$$ 2, 1)

Therefore, angle between the lines is given as :

cos ($$\theta $$) = $${{\left( { - 2} \right)\left( 1 \right) + \left( { - 1} \right).\left( { - 2} \right) + \left( 1 \right)\left( 1 \right)} \over {\sqrt 6 .\sqrt 6 }}$$

$$ \Rightarrow $$ cos ($$\theta $$) = $${1 \over 6}$$ $$ \Rightarrow $$ $$\theta $$ =cos^{$$-$$1} $$\left( {{1 \over 6}} \right)$$

l + 3m + 5n = 0

and 5$$l$$m $$-$$ 2mn + 6n$$l$$ = 0

From eq. (1) we have

$$l$$ = $$-$$ 3m $$-$$ 5n

Put the value of $$l$$ in eq. (2), we get ;

5 ($$-$$3m $$-$$5n) m $$-$$ 2mn + 6n ($$-$$ 3m $$-$$ 5n) = 0

$$ \Rightarrow $$ 15m

$$ \Rightarrow $$ m

$$ \Rightarrow $$ m

$$ \Rightarrow $$ $$\,\,\,$$ (m + n) (m + 2n) = 0

$$ \therefore $$ m = $$-$$ n or m = $$-$$ 2n

For m = $$-n, $$ $$l$$ = $$-$$ 2n

And for m = $$-$$ 2n, $$l$$ = n

$$ \therefore $$ ($$l$$, m, n) = ($$-$$2n, $$-$$n, n) Or ($$l$$, m, n) = (n, $$-$$ 2n, n)

$$ \Rightarrow $$ ($$l$$, m, n) = ($$-$$2, $$-$$1, 1) Or ($$l$$, m, n) = (1, $$-$$ 2, 1)

Therefore, angle between the lines is given as :

cos ($$\theta $$) = $${{\left( { - 2} \right)\left( 1 \right) + \left( { - 1} \right).\left( { - 2} \right) + \left( 1 \right)\left( 1 \right)} \over {\sqrt 6 .\sqrt 6 }}$$

$$ \Rightarrow $$ cos ($$\theta $$) = $${1 \over 6}$$ $$ \Rightarrow $$ $$\theta $$ =cos

2

A plane bisects the line segment joining the points (1, 2, 3) and ($$-$$ 3, 4, 5) at rigt angles. Then this plane also passes through the point :

A

($$-$$ 3, 2, 1)

B

(3, 2, 1)

C

($$-$$ 1, 2, 3)

D

(1, 2, $$-$$ 3)

Since the plane bisects the line joining the points (1, 2, 3) and ($$-$$3, 4, 5) then the plane passes through the midpoint of the line which is :

$$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$$ $$ \equiv $$ $$\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$$ $$ \equiv $$ ($$-$$1, 3, 4).

As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($$-$$ 3 $$-$$ 1, 4 $$-$$ 2, 5 $$-$$ 3) = ($$-$$ 4, 2, 2)

So the equation of the plane is : $$-$$ 4x + 2y + 2z = $$\lambda $$

As plane passes through ($$-$$ 1, 3, 4)

So, $$-$$ 4($$-$$ 1) + 2(3) + 2(4) = $$\lambda $$ $$ \Rightarrow $$ $$\lambda $$ = 18

Therefore, equation of plane is : $$-$$ 4x + 2y + 2z = 18

Now, only ($$-$$ 3, 2, 1) satiesfies the given plane as

$$-$$ 4($$-$$ 3) + 2(2) + 2(1) = 18

$$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$$ $$ \equiv $$ $$\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$$ $$ \equiv $$ ($$-$$1, 3, 4).

As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($$-$$ 3 $$-$$ 1, 4 $$-$$ 2, 5 $$-$$ 3) = ($$-$$ 4, 2, 2)

So the equation of the plane is : $$-$$ 4x + 2y + 2z = $$\lambda $$

As plane passes through ($$-$$ 1, 3, 4)

So, $$-$$ 4($$-$$ 1) + 2(3) + 2(4) = $$\lambda $$ $$ \Rightarrow $$ $$\lambda $$ = 18

Therefore, equation of plane is : $$-$$ 4x + 2y + 2z = 18

Now, only ($$-$$ 3, 2, 1) satiesfies the given plane as

$$-$$ 4($$-$$ 3) + 2(2) + 2(1) = 18

3

If the position vectors of the vertices A, B and C of a $$\Delta $$ ABC are respectively $$4\widehat i + 7\widehat j + 8\widehat k,$$ $$2\widehat i + 3\widehat j + 4\widehat k,$$ and $$2\widehat i + 5\widehat j + 7\widehat k,$$ then the position vectors of the point, where the bisector of $$\angle $$A meets BC is :

A

$${1 \over 2}\left( {4\widehat i + 8\widehat j + 11\widehat k} \right)$$

B

$${1 \over 3}\left( {6\widehat i + 11\widehat j + 15\widehat k} \right)$$

C

$${1 \over 3}\left( {6\widehat i + 13\widehat j + 18\widehat k} \right)$$

D

$${1 \over 4}\left( {8\widehat i + 14\widehat j + 19\widehat k} \right)$$

Suppose angular bisector of A meets BC at D(x, , z)

Using angular bisector theorem,

$${{AB} \over {AC}}$$ = $${{BD} \over {DC}}$$

$${{BD} \over {DC}}$$ = $${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$$

= $${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$$ = $${6 \over 3}$$ = 2

So, D(x, y, z) $$ \equiv $$ $$\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$$

$$\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$$

D(x, y, z) $$ \equiv $$ $$\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$$

Therefore, position vector of point p = $${1 \over 3}$$ (6i + 13j + 18k)

Using angular bisector theorem,

$${{AB} \over {AC}}$$ = $${{BD} \over {DC}}$$

$${{BD} \over {DC}}$$ = $${{\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {8 - 4} \right)}^2}} } \over {\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 5} \right)}^2} + {{\left( {8 - 7} \right)}^2}} }}$$

= $${{\sqrt {{2^2} + {4^2} + {4^2}} } \over {\sqrt {{2^2} + {2^2} + {1^2}} }}$$ = $${6 \over 3}$$ = 2

So, D(x, y, z) $$ \equiv $$ $$\left( {{{\left( 2 \right)\left( 2 \right) + \left( 1 \right)\left( 2 \right)} \over {2 + 1}},{{\left( 2 \right)\left( 5 \right) + \left( 1 \right)\left( 3 \right)} \over {2 + 1}}} \right.$$

$$\left. {{{\left( 2 \right)\left( 7 \right) + \left( 1 \right)\left( 4 \right)} \over {2 + 1}}} \right)$$

D(x, y, z) $$ \equiv $$ $$\left( {{6 \over 3},{{13} \over 3},{{18} \over 3}} \right)$$

Therefore, position vector of point p = $${1 \over 3}$$ (6i + 13j + 18k)

4

The sum of the intercepts on the coordinate axes of the plane passing through the point ($$-$$2, $$-2,$$ 2) and containing the line joining the points (1, $$-$$1, 2) and (1, 1, 1) is :

A

4

B

$$-$$ 4

C

$$-$$ 8

D

12

Equation of plane passing through three given points is :

$$\left| {\matrix{ {x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr {1 + 2} & { - 1 + 2} & {2 - 2} \cr {1 + 2} & {1 + 2} & {1 - 2} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr 3 & 1 & 0 \cr 3 & {30} & { - 1} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$ - x + 3y + 6z - 8 = 0$$

$$ \Rightarrow $$ $${x \over 8} - {{3y} \over 8} - {{6z} \over 8} + {8 \over 8} = 0$$

$$ \Rightarrow $$ $${x \over 8} - {y \over {{8 \over 3}}} - {z \over {{8 \over 6}}} = - 1$$

$$ \Rightarrow $$ $${x \over { - 8}} + {y \over {{8 \over 3}}} + {z \over {{8 \over 6}}} = 1$$

$$ \therefore $$ Sum of intercepts $$ = - 8 + {8 \over 3} + {8 \over 6} = - 4$$

$$\left| {\matrix{ {x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr {1 + 2} & { - 1 + 2} & {2 - 2} \cr {1 + 2} & {1 + 2} & {1 - 2} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr 3 & 1 & 0 \cr 3 & {30} & { - 1} \cr } } \right| = 0$$

$$ \Rightarrow $$ $$ - x + 3y + 6z - 8 = 0$$

$$ \Rightarrow $$ $${x \over 8} - {{3y} \over 8} - {{6z} \over 8} + {8 \over 8} = 0$$

$$ \Rightarrow $$ $${x \over 8} - {y \over {{8 \over 3}}} - {z \over {{8 \over 6}}} = - 1$$

$$ \Rightarrow $$ $${x \over { - 8}} + {y \over {{8 \over 3}}} + {z \over {{8 \over 6}}} = 1$$

$$ \therefore $$ Sum of intercepts $$ = - 8 + {8 \over 3} + {8 \over 6} = - 4$$

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