3D Geometry — Page 27 | JEE Mathematics

Q261

The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, – 1), having normal perpendicular to both the lines

{{x - 1} \over 1} = {{y + 2} \over { - 2}} = {{z - 4} \over 3}

and

{{x - 2} \over 2} = {{y + 1} \over { - 1}} = {{z + 7} \over { - 1}}

is :

A

{{10} \over {\sqrt {83} }}

B

{{5} \over {\sqrt {83} }}

C

{{10} \over {\sqrt {74} }}

D

{{20} \over {\sqrt {74} }}

Correct Answer

Option A

Solution

Let the plane be a(x $-$ 1) + b(y + 1) + c (z + 1) = 0 Normal vector

\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 1 & { - 2} & 3 \\ 2 & { - 1} & { - 1} \end{array} \right| = 5\widehat i + 7\widehat j + 3\widehat k

So plane is 5(x $-$ 1) + 7(y + 1) + 3(z + 1) = 0 $\Rightarrow$ 5x + 7y + 3z + 5 = 0 Distance of point (1, 3, $-$ 7) from the plane is

{{5 + 21 - 21 + 5} \over {\sqrt {25 + 49 + 9} }} = {{10} \over {\sqrt {83} }}

Q262

If the angle between the lines,

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

and

{{5 - x} \over { - 2}} = {{7y - 14} \over p} = {{z - 3} \over 4}\,\,

is

{\cos ^{ - 1}}\left( {{2 \over 3}} \right),

then p is equal to :

A

{7 \over 2}

B

{2 \over 7}

C

-

{7 \over 4}

D

-

{4 \over 7}

Correct Answer

Option A

Solution

Let $\theta$ be the angle between the two lines Here direction cosines of

{x \over 2}

=

{y \over 2}

=

{z \over 1}

are 2, 2, 1 Also second line can be written as :

{{x - 5} \over 2} = {{y - 2} \over {{P \over 7}}} = {{z - 3} \over 4}

$\therefore$ its direction cosines are 2,

{{P \over 7}}

, 4 Also, cos $\theta$ =

{2 \over 3}

(Given) $\because$ cos $\theta$

= \left| {{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \over {\sqrt {a_1^2 + b_1^2 + c_1^2\sqrt {a_2^2 + b_2^2 + c_2^2} } }}} \right|

$\Rightarrow$

{2 \over 3}

= \left| {{{\left( {2 \times 2} \right) + \left( {2 \times {P \over 7}} \right) + \left( {1 \times 4} \right)} \over {\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}} \right|

= {{4 + {{2P} \over 7} + 4} \over {3 \times \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}

$\Rightarrow$

{\left( {4 + {P \over 7}} \right)^2} = 20 + {{{P^2}} \over {49}}

$\Rightarrow$ 16 +

{{8P} \over 7} + {{{P^2}} \over {49}}

= 20 +

{{{P^2}} \over {49}}

$\Rightarrow$

{{8P} \over 7} = 4

$\Rightarrow$

P = {7 \over 2}

Q263

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

Correct Answer

Option B

Solution

Equation of plane passing through three given points is :

\left| \begin{array}{lll}{x - {x_1}} & {y - {y_1}} & {z - {z_1}} \\ {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \\ {{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \end{array} \right| = 0

$\Rightarrow$

\left| \begin{array}{lll}{x + 2} & {y + 2} & {z - 2} \\ {1 + 2} & { - 1 + 2} & {2 - 2} \\ {1 + 2} & {1 + 2} & {1 - 2} \end{array} \right| = 0

$\Rightarrow$

\left| \begin{array}{lll}{x + 2} & {y + 2} & {z - 2} \\ 3 & 1 & 0 \\ 3 & {30} & { - 1} \end{array} \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

Q264

An angle between the plane, x + y + z = 5 and the line of intersection of the planes, 3x + 4y + z

-

1 = 0 and 5x + 8y + 2z + 14 =0, is :

A

{\sin ^{ - 1}}\left( {\sqrt {{{3}/{{17}}}} } \right)

B

{\cos ^{ - 1}}\left( {\sqrt {{{3}/{{17}}}} } \right)

C

{\cos ^{ - 1}}\left( {{{3}/{{17}}}} \right)

D

{\sin ^{ - 1}}\left( {{{3}/{{17}}}} \right)

Correct Answer

Option A

Solution

Normal to

3x + 4y + z = 1

is

3\widehat i + 4\widehat j + \widehat k

Normal to

5x + 8y + 2z =

- 14

is

5\widehat i + 8\widehat j + 2\widehat k

The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors.

Therefore the direction ratios of the line is

- \widehat j + 4\widehat k

Hence the angle between the plane x + y + z + 5 = 0 and the intersection line is

{\sin ^{ - 1}}\left( {{{ - 1 + 4} \over {\sqrt {17} \sqrt 3 }}} \right) = {\sin ^{ - 1}}\left( {\sqrt {{3 \over {17}}} } \right)

Q265

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 :

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)

Correct Answer

Option C

Solution

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 $\Rightarrow$ 15m2 + 45mn + 30n2 = 0 $\Rightarrow$ m2 + 3mn + 2n2 = 0 $\Rightarrow$ m2 + 2mn + mn + 2n2 = 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)

Q266

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

Correct Answer

Option C

Solution

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}} }}

$\Rightarrow$ a2 $=$ $-$ 2bc and b $=$ $-$ c we get a2 $=$ 2c2 $\Rightarrow$ a $=$ $\pm$

\sqrt 2

c $\Rightarrow$ direction ratio (a, b, c) = (

\sqrt 2

, $-$ 1, 1) or (

\sqrt 2

, 1, $-$ 1)

Q267

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)

Correct Answer

Option A

Solution

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

Q268

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

Correct Answer

Option B

Solution

Normal vector of plane

= \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 2 & { - 5} & 0 \\ 4 & { - 4} & 5 \end{array} \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

Q269

If an angle between the line,

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

and the plane,

x - 2y - kz = 3

is

{\cos ^{ - 1}}\left( {{{2\sqrt 2 } \over 3}} \right),

then a value of k is :

A

\sqrt {{3 \over 5}}

B

- {5 \over 2}

C

- {3 \over 2}

D

\sqrt {{5 \over 3}}

Correct Answer

Option D

Solution

DR's of line are 2, 1, $-$ 2 normal vector of plane is

\widehat i

$-$ 2

\widehat j

$-$ k

\widehat k

sin $\alpha$ =

{{\left( {2\widehat i + \widehat j - 2\widehat k} \right).\left( {\widehat i - 2\widehat j - k\widehat k} \right)} \over {3\sqrt {1 + 4 + {k^2}} }}

sin $\alpha$ =

{{2k} \over {3\sqrt {{k^2} + 5} }}

. . . . . . (1) cos $\alpha$ =

{{2\sqrt 2 } \over 3}

. . . . .. . (2) (1)2 + (2)2 = 1 $\Rightarrow$ k2 =

{5 \over 3}

Q270

The equation of the line passing through (–4, 3, 1), parallel to the plane x + 2y – z – 5 = 0 and intersecting the line

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

is :

A

{{x + 4} \over 3} = {{y - 3} \over {-1}} = {{z - 1} \over 1}

B

{{x + 4} \over 1} = {{y - 3} \over {1}} = {{z - 1} \over 3}

C

{{x + 4} \over -1} = {{y - 3} \over {1}} = {{z - 1} \over 1}

D

{{x - 4} \over 2} = {{y + 3} \over {1}} = {{z + 1} \over 4}

Correct Answer

Option A

Solution

The line L is parallel to the plane P and intersect with line 4 at point R.

Let the coordinate of point R is, (x1, y1, z1) and it passes through L2.

{{{x_1} + 1} \over { - 3}} = {{{y_1} - 3} \over 2} = {{{z_1} - 2} \over { - 1}} = t

$\therefore$ x1 = $-$ 1 $-$ 3t, y1 = 3 + 2t, z1 = 2 $-$ t

\overrightarrow {AR} = \left( {3 - 3t} \right)\widehat i + \left( {2t} \right)\widehat j + \left( {1 - t} \right)\widehat k

\overrightarrow n = \widehat i + 2\widehat j - \widehat k

As

\overrightarrow {AR}

and

\overrightarrow n

are perpendicular to each other, So

\overrightarrow {AR}

$\cdot$

\overrightarrow n

= 0 $\Rightarrow$ (3 $-$ 3t) 1 + (2t)2 + (1 $-$ t) ( $-$ 1) = 0 $\Rightarrow$ 3 $-$ 3t + 4t $-$ 1 + t = 0 $\Rightarrow$ 2 + 2t = 0 $\Rightarrow$ t = $-$ 1 $\therefore$ point R = (2, 1, 3) $\therefore$ DR of line L is = (2 $-$ ( $-$ 4),

1

$-$ 3, 3 $-$ 1) = (6, $-$ 2, 2) $\therefore$ Equation of line is

{{x + 4} \over 6} = {{y - 3} \over { - 2}} = {{z - 1} \over 2}

or

{{x + 4} \over 3} = {{y - 3} \over { - 1}} = {{z - 1} \over 1}