3D Geometry — Page 8 | JEE Mathematics

Q71

The vector equation of the plane passing through the intersection of the planes

\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1

and

\overrightarrow r .\left( {\widehat i - 2\widehat j} \right) = - 2

, and the point (1, 0, 2) is :

A

\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = {7 \over 3}

B

\overrightarrow r .\left( {\widehat i + 7\widehat j + 3\widehat k} \right) = 7

C

\overrightarrow r .\left( {3\widehat i + 7\widehat j + 3\widehat k} \right) = 7

D

\overrightarrow r .\left( {\widehat i - 7\widehat j + 3\widehat k} \right) = {7 \over 3}

Correct Answer

Option B

Solution

Given, point (1, 0, 2) Equation of plane =

\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) = 1

and

\overrightarrow r\,.\,(\widehat i - 2\widehat j) = - 2

Equation of plane passing through the intersection of given planes is

[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \lambda [\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0

$\because$ This plane passes through point (1, 0, 2) i.e., vector

(\widehat i + 2\widehat k)

$\therefore$

[(\widehat i + 2\widehat k)\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \lambda [(\widehat i + 2\widehat k)\,.\,(\widehat i - 2\widehat j) + 2] = 0

\Rightarrow (3 - 1) + \lambda (1 + 2) = 0

\Rightarrow 2 + \lambda \times 3 = 0

\Rightarrow \lambda = - 2/3

Hence, equation of required plane is

[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] + \left( {{{ - 2} \over 3}} \right)[\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0

$\Rightarrow$

3[\overrightarrow r\,.\,(\widehat i + \widehat j + \widehat k) - 1] - 2[\overrightarrow r\,.\,(\widehat i - 2\widehat j) + 2] = 0

$\Rightarrow$

\overrightarrow r\,.\,(\widehat i + 7\widehat j + 3\widehat k) = 7

Q72

The equation of the line through the point (0, 1, 2) and perpendicular to the line

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

is :

A

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

B

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

C

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

D

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

Correct Answer

Option C

Solution

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

Any point on this line

(2\lambda + 1,3\lambda - 1, - 2\lambda + 1)

Direction ratio of given line

(2,3, - 2)

Direction ratio of line to be found

(2\lambda + 1,3\lambda - 2, - 2\lambda - 1)

$\therefore$

{\overrightarrow d _1}\,.\,{\overrightarrow d _2} = 0

$\Rightarrow$

\lambda = 2/17

Direction ratio of line

(21, - 28, - 21) \equiv (3, - 4, - 3) \equiv ( - 3,4,3)

Q73

Consider the line L given by the equation

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

. Let Q be the mirror image of the point (2, 3,

-

1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?

A (

-

1, 1, 2)

B (1, 1, 1)

C (1, 1, 2)

D (1, 2, 2)

Correct Answer

Option D

Solution

Plane p is

{ \bot ^r}

to line

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

& passes through pt. (2, 3) equation of plane p 2(x $-$ 2) + 1(y $-$ 3) + 1 (z + 1) = 0 2x + y + z $-$ 6 = 0 Point (1, 2, 2) satisfies above equation

Q74

A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2,

-

1, 1), then the projection of

\overrightarrow {OP}

on this plane is of length :

A

\sqrt {{2 \over 7}}

B

\sqrt {{2 \over 5}}

C

\sqrt {{2 \over 3}}

D

\sqrt {{2 \over 11}}

Correct Answer

Option D

Solution

A(1, 2, 3), B(2, 3, 1), C(2, 4, 2), O(0, 0, 0) Equation of plane passing through A, B, C will be

\left| \begin{array}{lll}{x - 1} & {y - 2} & {z - 3} \\ {2 - 1} & {3 - 2} & {1 - 3} \\ {2 - 1} & {4 - 2} & {2 - 3} \end{array} \right| = 0

\Rightarrow \left| \begin{array}{lll}{x - 1} & {y - 2} & {z - 3} \\ 1 & 1 & { - 2} \\ 1 & 2 & { - 1} \end{array} \right| = 0

\Rightarrow (x - 1)( - 1 + 4) - (y - 2)( - 1 + 2) + (z - 3)(2 - 1) = 0

\Rightarrow (x - 1)(3) - (y - 2)(1) + (z - 3)(1) = 0

\Rightarrow 3x - 3 - y + 2 + z - 3 = 0

\Rightarrow 3x - y + z - 4 = 0

, is the required plane. Now, O(0, 0, 0) & P(2, $-$ 1, 1) Plane is

3x - y + z - 4 = 0

O' & P' are foot of perpendiculars. For O'

{{x - 0} \over 3} = {{y - 0} \over { - 1}} = {{z - 0} \over 1} = {{ - (0 - 0 + 0 - 4)} \over {9 + 1 + 1}}

{x \over 3} = {y \over { - 1}} = {z \over 1} = {4 \over {11}}

\Rightarrow O'\left( {{{12} \over {11}},{{ - 4} \over {11}},{4 \over {11}}} \right)

for P'

{{x - 2} \over 3} = {{y + 1} \over { - 1}} = {{z - 1} \over 1} = {{ - (3(2) - ( - 1) + 1 - 4)} \over {9 + 1 + 1}}

{{x - 2} \over 3} = {{y + 1} \over { - 1}} = {{z - 1} \over 1} = \left( {{{ - 4} \over {11}}} \right)

P'\left( {{{ - 12} \over {11}} + 2,{4 \over {11}} - 1,{{ - 4} \over {11}} + 1} \right)

\Rightarrow P'\left( {{{10} \over {11}},{{ - 7} \over {11}},{7 \over {11}}} \right)

O'P' = \sqrt {{{\left( {{{10} \over {11}} - {{12} \over {11}}} \right)}^2} + {{\left( {{{ - 7} \over {11}} + {4 \over {11}}} \right)}^2} + {{\left( {{7 \over {11}} - {4 \over {11}}} \right)}^2}}

\Rightarrow O'P' = {1 \over {11}}\sqrt {4 + 9 + 9}

\Rightarrow O'P' = {{\sqrt {22} } \over {11}}

\Rightarrow O'P' = {{\sqrt 2 \times \sqrt {11} } \over {11}}

\Rightarrow O'P' = \sqrt {{2 \over {11}}}

Q75

Consider the three planes P1 : 3x + 15y + 21z = 9, P2 : x

-

3y

-

z = 5, and P3 : 2x + 10y + 14z = 5 Then, which one of the following is true?

A P1 and P2 are parallel.

B P1, P2 and P3 all are parallel.

C P1 and P3 are parallel.

D P2 and P3 are parallel.

Correct Answer

Option C

Solution

P1 : 3x + 15y + 21z = 9, P2 : x $-$ 3y $-$ z = 5 P3 : x + 5y + 7z = 5/2 $\therefore$ P1 and P3 are parallel.

Q76

If (1, 5, 35), (7, 5, 5), (1,

\lambda

, 7) and (2

\lambda

, 1, 2) are coplanar, then the sum of all possible values of

\lambda

is :

A

- {{44} \over 5}

B

- {{39} \over 5}

C

{{44} \over 5}

D

{{39} \over 5}

Correct Answer

Option C

Solution

A(1, 5, 35), B(7, 5, 5), C(1, $\lambda$ , 7), D(2 $\lambda$ , 1, 2)

\overrightarrow {AB}

= 6

\widehat i

$-$ 30

\widehat k

,

\overrightarrow {BC}

= $-$ 6

\widehat i

( $\lambda$ $-$ 5)

\widehat j

+ 2

\widehat k

,

\overrightarrow {CD}

= (2 $\lambda$ $-$ 1)

\widehat i

+ (1 $-$ $\lambda$ )

\widehat j

$-$ 5

\widehat k

Points are coplanar

\Rightarrow 0 = \left| \begin{array}{lll}6 & 0 & { - 30} \\ { - 6} & {\lambda - 5} & 2 \\ {2\lambda - 1} & {1 - \lambda } & { - 5} \end{array} \right|

= 6( $-$ 5 $\lambda$ + 25 $-$ 2 + 2 $\lambda$ ) $-$ 30( $-$ 6 + 6 $\lambda$ $-$ (2 $\lambda$ 2 $-$ $\lambda$ $-$ 10 $\lambda$ + 5)) = 6( $-$ 3 $\lambda$ + 23) $-$ 30( $-$ 2 $\lambda$ 2 + 11 $\lambda$ $-$ 5 $-$ 6 + 6 $\lambda$ ) = 6( $-$ 3 $\lambda$ + 23) $-$ 30( $-$ 2 $\lambda$ 2 + 17 $\lambda$ $-$ 11) = 6( $-$ 3 $\lambda$ + 23 + 10 $\lambda$ 2 $-$ 85 $\lambda$ + 55) = 6(10 $\lambda$ 2 $-$ 88 $\lambda$ + 78) = 12(5 $\lambda$ 2 $-$ 44 $\lambda$ + 39) $\Rightarrow$ 0 = 12(5 $\lambda$ 2 $-$ 44 $\lambda$ + 39) $\Rightarrow$ 5 $\lambda$ 2 $-$ 44 $\lambda$ + 39 = 0 this quadratic equation has two values $\lambda$ 1 and $\lambda$ 2 $\therefore$ $\lambda$ 1 + $\lambda$ 2 =

{{44} \over 5}

Q77

If the mirror image of the point (1, 3, 5) with respect to the plane 4x

-

5y + 2z = 8 is (

\alpha

,

\beta

,

\gamma

), then 5(

\alpha

+

\beta

+

\gamma

) equals :

A 39

B 41

C 47

D 43

Correct Answer

Option C

Solution

Image of (1, 3, 5) in the plane 4x $-$ 5y + 2z = 8 is ( $\alpha$ , $\beta$ , $\gamma$ )

\Rightarrow {{\alpha - 1} \over 4} = {{\beta - 3} \over { - 5}} = {{\gamma - 5} \over 2} = - {{(4(1) - 5(3) + 2(5) - 8)} \over {{4^2} + {5^2} + {2^2}}} = {2 \over 5}

$\therefore$

\alpha = 1 + 4\left( {{2 \over 5}} \right) = {{13} \over 5}

\beta = 3 - 5\left( {{2 \over 5}} \right) = 1 = {5 \over 5}

\gamma = 5 + 2\left( {{2 \over 5}} \right) = {{29} \over 5}

Thus,

5(\alpha + \beta + \gamma ) = 5\left( {{{13} \over 5} + {5 \over 5} + {{29} \over 5}} \right) = 47

Q78

Let L be a line obtained from the intersection of two planes x + 2y + z = 6 and y + 2z = 4. If point P(

\alpha

,

\beta

,

\gamma

) is the foot of perpendicular from (3, 2, 1) on L, then the value of 21(

\alpha

+

\beta

+

\gamma

) equals :

A 102

B 142

C 136

D 68

Correct Answer

Option A

Solution

Dr's of line

\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{array} \right| = 3\widehat i - 2\widehat j + \widehat k

Dr/s : - (3, $-$ 2, 1) Points on the line ( $-$ 2, 4, 0) Equation of the line

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

Dr's of PQ :

3\lambda - 5, - 2\lambda + 2,\lambda - 1

Dr's of y lines are (3, $-$ 2, 1) Since

PQ \bot

line

3(3\lambda - 5) - 2( - 2\lambda + 2) + 1(\lambda - 1) = 0

\lambda = {{10} \over 7}

P\left( {{{16} \over 7},{8 \over 7},{{10} \over 7}} \right)

21(\alpha + \beta + \gamma ) = 21\left( {{{34} \over 7}} \right) = 102

Q79

Let the position vectors of two points P and Q be 3

\widehat i

-

\widehat j

+ 2

\widehat k

and

\widehat i

+ 2

\widehat j

-

4

\widehat k

, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4,

-

1, 2) and (

-

2, 1,

-

2), respectively. Let lines PR and QS intersect at T. If the vector

\overrightarrow {TA}

is perpendicular to both

\overrightarrow {PR}

and

\overrightarrow {QS}

and the length of vector

\overrightarrow {TA}

is

\sqrt 5

units, then the modulus of a position vector of A is :

A

\sqrt {171}

B

\sqrt {227}

C

\sqrt {482}

D

\sqrt {5}

Correct Answer

Option A

Solution

\overrightarrow p = 3\widehat i - \widehat j + 2\widehat k

&

\overrightarrow Q = \widehat i + 2\widehat j - 4\widehat k

{\overrightarrow v _{PR}} = (4, - 1,2)

&

{\overrightarrow v _{QS}} = ( - 2,1, - 2)

{L_{PR}}:\overrightarrow r = (3\widehat i - \widehat j + 2\widehat k) + \lambda (4, - 1,2)

{L_{QS}}:\overrightarrow r = (\widehat i + 2\widehat j - 4\widehat k) + \mu ( - 2,1, - 2)

Now T on PR =

\left\langle {3 + 4\lambda , - 1 - \lambda ,2 + 2\lambda } \right\rangle

Similarly T on QS = (1 $-$ 2 $\mu$ , 2 + $\mu$ , $-$ 4 $-$ 2 $\mu$ ) For $\lambda$ & $\mu$ :

\left. \begin{array}{ll}3 + 4\lambda = 1 - 2\mu \Rightarrow \mu + 2\lambda = - 1 \\ - 1 - \lambda = 2 + \mu \Rightarrow \mu + \lambda = - 3 \end{array} \right\}\begin{array}{ll}{\lambda = 2} \\ {\mu = - 5} \end{array}

$\Rightarrow$ T : (11, $-$ 3, 6) D.R. of TA =

{\overrightarrow v _{QS}}

$\times$

{\overrightarrow v _{PR}}

=

\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ { - 2} & 1 & { - 2} \\ 4 & { - 1} & 2 \end{array} \right| = 0\widehat i - 4\widehat j - 2\widehat k

{L_{TA}}:\overrightarrow r = \left( {11\widehat i - 3\widehat j + 6\widehat k} \right) + \lambda ( - 4\widehat j - 2\widehat k)

Now A = (11, $-$ 3 $-$ 4 $\lambda$ , 6 $-$ 2 $\lambda$ ) Given, TA =

\sqrt 5

{( - 3 + 4\lambda + 3)^2} + {(6 + 2\lambda - 6)^2} = 5

16{\lambda ^2} + 4{\lambda ^2} = 5

20 $\lambda$ 2 = 5 $\lambda$ = $\pm$

{1 \over 2}

\begin{array}{lll}{A:(11, - 3 - 2,6 - 1)} & ; & {A:(11, - 3 + 2,6 + 1)} \\ {|A|\, = \sqrt {121 + 25 + 25} } & ; & {|A|\, = \sqrt {121 + 1 + 49} } \\ { = \sqrt {171} } & ; & {\sqrt {171} } \end{array}

Q80

Let P be a plane lx + my + nz = 0 containing the line,

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

. If plane P divides the line segment AB joining points A(

-

3,

-

6, 1) and B(2, 4,

-

3) in ratio k : 1 then the value of k is equal to :

A 2

B 3

C 1.5

D 4

Correct Answer

Option A

Solution

Line lies on plane

- l + 2m + 3n = 0

..... (1) Point on line (1, $-$ 4, $-$ 2) lies on plane

l - 4m - 2n = 0

.... (2) from (1) & (2)

- 2m + n = 0 \Rightarrow 2m = n

l = 3n + 2m \Rightarrow l = 4n

l:m:n::4n:{n \over 2}:n

l:m:n::8n:n:2n

l:m:n::8:1:2

Now equation of plane is 8x + y + 2z = 0 R divide AB is ratio k : 1

R:\left( {{{ - 3 + 2k} \over {k + 1}},{{ - 6 + 4k} \over {k + 1}},{{1 - 3k} \over {k + 1}}} \right)

lies on plane

8\left( {{{ - 3 + 2k} \over {k + 1}}} \right) + \left( {{{ - 6 + 4k} \over {k + 1}}} \right) + 2\left( {{{1 - 3k} \over {k + 1}}} \right) = 0

- 24 + 16k - 6 + 4k + 2 - 6k = 0

- 28 + 14k = 0

k = 2