3D Geometry — Page 6 | JEE Mathematics

Q51

The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines

\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right)

and

\overrightarrow r = \left( {\widehat i + \widehat j} \right) + \mu \left( { - \widehat i + \widehat j - 2\widehat k} \right)

is :

A

{1 \over 3}

B

{1 \over {\sqrt 3 }}

C 3

D

{\sqrt 3 }

Correct Answer

Option D

Solution

Vector of the plane is

\left| \begin{array}{lll}{\hat i} & {\hat j} & {\hat k} \\ 1 & 2 & { - 1} \\ { - 1} & 1 & { - 2} \end{array} \right| = - 3\hat i + 3\hat j + 3\hat k

Now equation of plane is

- 3x + 3y + 3z = c

(1, 1, 0) will satisfy the plane

\Rightarrow - 3 + 3 + 0 = c

$\Rightarrow$ c = 0

- 3x + 3y + 3z = 0

distance from (2, 1, 4) is

\Rightarrow \left| {{{ - 6 + 3 + 12} \over {\sqrt {27} }}} \right| = \left| {{9 \over {3\sqrt 3 }}} \right| = \sqrt 3 \,\,units

Q52

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)

Correct Answer

Option C

Solution

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)

Q53

A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(–1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is :

A cos

-

1

\left( {{{17} \over {31}}} \right)

B cos

-

1

\left( {{{9} \over {35}}} \right)

C cos

-

1

\left( {{{19} \over {35}}} \right)

D cos

-

1

\left( {{7 \over {31}}} \right)

Correct Answer

Option C

Solution

\overrightarrow {OP} \times \overrightarrow {OQ} = \left( {\widehat i + 2\widehat j + \widehat k} \right) \times \left( {2\widehat i + \widehat j + 3\widehat k} \right)

= 5

\widehat i

$-$

\widehat j

$-$ 3

\widehat k

\overrightarrow {PQ} \times \overrightarrow {PR} = \left( {\widehat i - \widehat j + 2\widehat k} \right) \times \left( { - 2\widehat i - \widehat j + \widehat k} \right)

=

\widehat i

$-$ 5

\widehat j

$-$ 3

\widehat k

cos $\theta$ =

{{5 + 5 + 9} \over {{{\left( {\sqrt {25 + 9 + 1} } \right)}^2}}} = {{19} \over {35}}

Q54

The length of the perpendicular from the point (2, –1, 4) on the straight line,

{{x + 3} \over {10}}

=

{{y - 2} \over {-7}}

=

{{z} \over {1}}

is :

A less than 2

B greater than 4

C greater than 2 but less than 3

D greater than 3 but less than 4

Correct Answer

Option D

Solution

Let

{{x + 3} \over {10}}

=

{{y - 2} \over {-7}}

=

{{z} \over {1}}

= $\lambda$ $\therefore$ General point Q(10 $\lambda$ - 3, -7 $\lambda$ + 2, $\lambda$ ).

\overrightarrow {PQ}

= (10 $\lambda$ - 5)

\widehat i

+ (3 - 7 $\lambda$ )

\widehat j

+ ( $\lambda$ - 4)

\widehat k

The vector parrallel to the given line is = 10

\widehat i

- 7

\widehat j

+

\widehat k

As

\overrightarrow {PQ}

and (10

\widehat i

- 7

\widehat j

+

\widehat k

) are perpendicular to the each other.

\overrightarrow {PQ}

. (10

\widehat i

- 7

\widehat j

+

\widehat k

) = 0 $\Rightarrow$ (10 $\lambda$ - 5)10 + (3 - 7 $\lambda$ )(-7) + ( $\lambda$ - 4)(1) = 0 $\Rightarrow$ 100 $\lambda$ – 50 + 49 $\lambda$ – 21 + $\lambda$ – 4 = 0 $\Rightarrow$ 150 $\lambda$ = 75 $\Rightarrow$ $\lambda$ =

{1 \over 2}

$\therefore$ The point Q = (2,

- {3 \over 2}

,

{1 \over 2}

) $\therefore$ Length of perpendicular (PQ) =

\sqrt {0 + {1 \over 4} + {{49} \over 4}}

=

\sqrt {{{50} \over 4}} = {5 \over {\sqrt 2 }}

= 3.53

Q55

The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ?

A (2, 1, 3)

B (4,

-

1, 2)

C (4, 1,

-

2)

D (

-

2, 3, 5)

Correct Answer

Option C

Solution

p : 3(x $-$ 0) + 5(y $-$ 2) + 1 (z $-$ 5) = 0 3x + 5y + z = 15 On checking all the options, the option (4, 1, − 2) satisfy the equation of plane.

Q56

The plane which bisects the line joining, the points (4, –2, 3) and (2, 4, –1) at right angles also passes through the point :

A (4, 0, 1)

B (0, –1, 1)

C (0, 1, –1)

D (4, 0, –1)

Correct Answer

Option D

Solution

Direction ratios of normal to plane are < 2, –6, 4 > Also plane passes through (3, 1, 1) $\therefore$ Equation of plane 2(x–3)–6(y–1)+4(z–1) = 0 $\Rightarrow$ x – 3y + 2z = 2 By checking all options we can see this equation passes through (4, 0, –1)

Q57

The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line

{x \over 2} = {y \over 3} = {z \over { - 6}}

is :

A 7

B 1

C

{1 \over 7}

D

{7 \over 5}

Correct Answer

Option B

Solution

Equation of line parallel to

{x \over 2} = {y \over 3} = {z \over { - 6}}

passes through

(1, - 2,3)

is

{{x - 1} \over 2} = {{y + 2} \over 3} = {{z - 3} \over { - 6}} = r

x = 2r + 1

y = 3r - 2

,

z = - 6r + 3

A point on whole line = (2r + 1, 3r – 2, – 6r + 3). This point lies on plane x – y + 2 = 5 so,

2r + 1 - 3r + 2 - 6r + 3 = 5

$\Rightarrow$

r = {1 \over 7}

$\therefore$

x = {9 \over 7}

,

y = {{ - 11} \over 7}

,

z = {{15} \over 7}

Distance is =

\sqrt {{{\left( {{9 \over 7} - 1} \right)}^2} + {{\left( {2 - {{11} \over 7}} \right)}^2} + {{\left( {3 - {{15} \over 7}} \right)}^2}}

= \sqrt {{{\left( {{2 \over 7}} \right)}^2} + {{\left( {{3 \over 7}} \right)}^2} + {{\left( {{6 \over 7}} \right)}^2}}

= {1 \over 7}\sqrt {4 + 9 + 36}

= 1

Q58

If (a, b, c) is the image of the point (1, 2, -3) in the line

{{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}

, then a + b + c is :

A 1

B 2

C 3

D -1

Correct Answer

Option B

Solution

Equation of line :

{{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}

= $\lambda$ (Assume) A point on line L is = R(2 $\lambda$ – 1, –2 $\lambda$ + 3, – $\lambda$ ) DR's of PR = < 2 $\lambda$ – 2, –2 $\lambda$ + 1, – $\lambda$ + 3 > $\because$ PR is perpendicular to line L $\therefore$ 2(2 $\lambda$ –2) –2 (–2 $\lambda$ + 1) –1 (– $\lambda$ + 3) = 0 $\Rightarrow$ 4 $\lambda$ – 4 + 4 $\lambda$ – 2 + $\lambda$ – 3 = 0 $\Rightarrow$ 9 $\lambda$ – 9 = 0 $\Rightarrow$ $\lambda$ = 1 $\therefore$ Coordinate of foot of perpendicular = R = (1, 1, –1) As R is the midpoint of line PQ, so

{{a + 1} \over 2} = 1

$\Rightarrow$ a = 1

{{b + 2} \over 2} = 1

$\Rightarrow$ b = 0

{{c - 3} \over 2} = - 1

$\Rightarrow$ c = 1 $\therefore$ a + b + c = 2

Q59

If for some

\alpha

\in

R, the lines L1 :

{{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}

and L2 :

{{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}

are coplanar, then the line L2 passes through the point :

A (10, 2, 2)

B (2, –10, –2)

C (10, –2, –2)

D (–2, 10, 2)

Correct Answer

Option B

Solution

L1 :

{{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}

and L2 :

{{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}

are coplanar. $\therefore$

\left| \begin{array}{lll}1 & 3 & 2 \\ 2 & { - 1} & 1 \\ \alpha & {5 - \alpha } & 1 \end{array} \right|

= 0 $\Rightarrow$ –1(–1 + $\alpha$ - 5) + 3(2 - $\alpha$ ) - 2(10 - 2 $\alpha$ + $\alpha$ ) = 0 $\Rightarrow$ 6 - $\alpha$ + 6 - 3 $\alpha$ + 2 $\alpha$ - 20 = 0 $\Rightarrow$ –8 –2 $\alpha$ = 0 $\Rightarrow$ $\alpha$ = -4 $\therefore$ Equation of L2 :

{{x + 2} \over { - 4}} = {{y + 1} \over 9} = {{z + 1} \over 1}

Check options (2, –10, –2) lies on L2.

Q60

The shortest distance between the lines

{{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}

and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :

A 1

B

{1 \over 2}

C

{1 \over {\sqrt 2 }}

D

{1 \over {\sqrt 3 }}

Correct Answer

Option D

Solution

Plane through line of intersection is x + y + z + 1 + $\lambda$ (2x –y + z + 3) = 0 It should be parallel to given line

{{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}

$\therefore$ 0(1 + 2 $\lambda$ ) - 1(1 - $\lambda$ ) + 1(1 + $\lambda$ ) = 0 $\Rightarrow$ $\lambda$ = 0 $\therefore$ Required Plane : x + y + z + 1 = 0 Shortest distance of (1, –1, 0) from this plane =

{{\left| {1 - 1 + 0 + 1} \right|} \over {\sqrt {{1^2} + {1^2} + {1^2}} }}

=

{1 \over {\sqrt 3 }}