3D Geometry — Page 14 | JEE Mathematics

Q131

If the length of the perpendicular drawn from the point

P(a, 4,2)

, a

>0

on the line

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

is

2 \sqrt{6}

units and

Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)

is the image of the point P in this line, then

\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}

is equal to :

A 7

B 8

C 12

D 14

Correct Answer

Option B

Solution

$\because$ PR is perpendicular to given line, so

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

\Rightarrow a = 7\lambda - 2

Now, $\because$

PR = 2\sqrt 6

\Rightarrow {( - 5\lambda + 1)^2} + {(3\lambda - 1)^2} + {(\lambda + 1)^2} = 24

\Rightarrow 5{\lambda ^2} - 2\lambda - 3 = 0 \Rightarrow \lambda = 1

or

- {3 \over 5}

$\because$

a > 0

so

\lambda = 1

and

a = 5

Now

\sum\limits_{i = 1}^3 {{\alpha _i} = 2}

(Sum of co-ordinate of R) $-$ (Sum of coordinates of P)

= 2(7) - 11 = 3

a + \sum\limits_{i = 1}^3 {{\alpha _i} = 5 + 3 = 8}

Q132

If the line of intersection of the planes

a x+b y=3

and

a x+b y+c z=0

, a

>0

makes an angle

30^{\circ}

with the plane

y-z+2=0

, then the direction cosines of the line are :

A

\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}, 0

B

\dfrac{1}{\sqrt{2}}, \pm \,\dfrac{1}{\sqrt{2}}, 0

C

\dfrac{1}{\sqrt{5}},-\dfrac{2}{\sqrt{5}}, 0

D

\dfrac{1}{2},-\dfrac{\sqrt{3}}{2}, 0

Correct Answer

Option B

Solution

{P_1}:ax + by + 0z = 3

, normal vector :

{\overrightarrow n _1} = (a,b,0)

{P_2}:ax + by + cz = 0

, normal vector :

{\overrightarrow n _2} = (a,b,c)

Vector parallel to the line of intersection

= {\overrightarrow n _1} \times {\overrightarrow n _2}

{\overrightarrow n _1} \times {\overrightarrow n _2} = (bc, - ac,0)

Vector normal to

0\,.\,x + y - z + 2 = 0

is

{\overrightarrow n _3} = (0,1, - 1)

Angle between line and plane is 30

^\circ

\Rightarrow \left| {{{0 - ac + 0} \over {\sqrt {{b^2}{c^2} + {c^2}{a^2}} \sqrt 2 }}} \right| = {1 \over 2}

\Rightarrow {a^2} = {b^2}

Hence,

{\overrightarrow n _1} \times {\overrightarrow n _2} = (ac, - ac,0)

Direction ratios

= \left( {{1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }},0} \right)

Q133

The foot of the perpendicular from a point on the circle

x^{2}+y^{2}=1, z=0

to the plane

2 x+3 y+z=6

lies on which one of the following curves?

A

(6 x+5 y-12)^{2}+4(3 x+7 y-8)^{2}=1, z=6-2 x-3 y

B

(5 x+6 y-12)^{2}+4(3 x+5 y-9)^{2}=1, z=6-2 x-3 y

C

(6 x+5 y-14)^{2}+9(3 x+5 y-7)^{2}=1, z=6-2 x-3 y

D

(5 x+6 y-14)^{2}+9(3 x+7 y-8)^{2}=1, z=6-2 x-3 y

Correct Answer

Option B

Solution

Any point on

{x^2} + {y^2} = 1

,

z = 0

is

p(\cos \theta ,\,\sin \theta ,\,0)

If foot of perpendicular of p on the plane

2x + 3y + z = 6

is

(h,k,l)

then

{{h - \cos \theta } \over 2} = {{k - \sin \theta } \over 3} = {{l - 0} \over 1}

= - \left( {{{2\cos \theta + 3\sin \theta + 0 - 6} \over {{2^2} + {3^2} + {1^2}}}} \right) = r

(let)

h = 2r + \cos \theta ,\,k = 3r + \sin \theta ,\,l = r

Hence,

h - 2l = \cos \theta

and

k - 3l = \sin \theta

Hence,

{(h - 2l)^2} + {(k - 3l)^2} = 1

When

l = 6 - 2h - 3k

Hence required locus is

{(x - 2(6 - 2x - 3y))^2} + {(y - 3(6 - 2x - 3y))^2} = 1

\Rightarrow {(5x + 6y - 12)^2} + 4{(3x + 5y - 9)^2} = 1,\,z = 6 - 2x - 3y

Q134

Let the lines

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

and

\dfrac{x+26}{-2}=\dfrac{y+18}{3}=\dfrac{z+28}{\lambda}

be coplanar and

\mathrm{P}

be the plane containing these two lines. Then which of the following points does NOT lie on P?

A

(0,-2,-2)

B

(-5,0,-1)

C

(3,-1,0)

D

(0,4,5)

Correct Answer

Option D

Solution

{L_1}:{{x - 1} \over \lambda } = {{y - 2} \over 1} = {{z - 3} \over 2}

, through a point

{\overrightarrow a _1} \equiv (1,2,3)

parallel to

{\overrightarrow b _1} \equiv (\lambda ,1,2)

{L_2}:{{x + 26} \over { - 2}} = {{y + 18} \over 3} = {{z + 28} \over \lambda }

through a point

{\overrightarrow a _2} = ( - 26, - 18, - 28)

parallel to

{\overrightarrow b _2} = ( - 2,3,1)

If lines are coplanar then,

({\overrightarrow a _2} - {\overrightarrow a _1})\,.\,{\overrightarrow b _1} \times {\overrightarrow b _2} = 0

\Rightarrow \left| \begin{array}{lll}{27} & {20} & {31} \\ \lambda & 1 & 2 \\ { - 2} & 3 & \lambda \end{array} \right| = 0 \Rightarrow \lambda = 3

Vector normal to the required plane

\overrightarrow n = {\overrightarrow b _1} \times {\overrightarrow b _2}

\Rightarrow \overrightarrow n = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 3 & 1 & 2 \\ { - 2} & 3 & 3 \end{array} \right| = - 3\widehat i - 13\widehat j + 11\widehat k

Equation of plane

\equiv ((x - 1),(y - 2),(z - 3))\,.\,( - 3, - 13,11) = 0

\Rightarrow 3x + 13y - 11z + 4 = 0

Checking the option gives (0, 4, 5) does not lie on the plane.

Q135

A plane P is parallel to two lines whose direction ratios are

-2,1,-3

and

-1,2,-2

and it contains the point

(2,2,-2)

. Let P intersect the co-ordinate axes at the points

\mathrm{A}, \mathrm{B}, \mathrm{C}

making the intercepts

\alpha, \beta, \gamma

. If

\mathrm{V}

is the volume of the tetrahedron

\mathrm{OABC}

, where

\mathrm{O}

is the origin, and

\mathrm{p}=\alpha+\beta+\gamma

, then the ordered pair

(\mathrm{V}, \mathrm{p})

is equal to :

A

(48,-13)

B

(24,-13)

C

(48,11)

D

(24,-5)

Correct Answer

Option B

Solution

Let

{\overrightarrow a _1} = ( - 2,1, - 3)

and

{\overrightarrow a _2} = ( - 1,2, - 2)

Vector normal to plane

\overline n = {\overrightarrow a _1} \times {\overrightarrow a _2}

\overline n = (4, - 1, - 3)

Plane through

(2,2, - 2)

and normal to

\overline n

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

\Rightarrow 4x - y - 3z = 12

\Rightarrow {x \over 3} + {y \over { - 12}} + {z \over { - 4}} = 1

Intercepts $\alpha$ , $\beta$ , $\gamma$ are

3, - 12, - 4

P = \alpha + \beta + \gamma = - 13

V = {1 \over 6} \times 3 \times 12 \times 4 = 24

Q136

If the foot of the perpendicular from the point

\mathrm{A}(-1,4,3)

on the plane

\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4

, is

\left(-2, \dfrac{7}{2}, \dfrac{3}{2}\right)

, then the distance of the point A from the plane P, measured parallel to a line with direction ratios

3,-1,-4

, is equal to :

A 1

B

\sqrt{26}

C 2

\sqrt{2}

D

\sqrt{14}

Correct Answer

Option B

Solution

\left( { - 2,{7 \over 2},{3 \over 2}} \right)

satisfies the plane

P:2x + my + nz = 4

- 4 + {{7m} \over 2} + {{3n} \over 2} = 4 \Rightarrow 7m + 3n = 16

...... (i) Line joining

A( - 1,\,4,\,3)

and

\left( { - 2,{7 \over 2},{3 \over 2}} \right)

is perpendicular to

P:2x + my + nz = 4

{1 \over 2} = {{{1 \over 2}} \over m} = {{{3 \over 2}} \over n} \Rightarrow m = 1

&

n = 3

Plane

P:2x + y + 3z = 4

Distance of P from

A( - 1,\,4,\,3)

parallel to the line

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

for point of intersection of P & L

2(3r - 1) + ( - r + 4) + 3( - 4r + 3) = 4 \Rightarrow r = 1

Point of intersection :

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

Required distance

= \sqrt {{3^2} + {1^2} + {4^2}} = \sqrt {26}

Q137

If

(2,3,9),(5,2,1),(1, \lambda, 8)

and

(\lambda, 2,3)

are coplanar, then the product of all possible values of

\lambda

is:

A

\dfrac{21}{2}

B

\dfrac{59}{8}

C

\dfrac{57}{8}

D

\dfrac{95}{8}

Correct Answer

Option D

Solution

$\because A(2,3,9), B(5,2,1), C(1, \lambda, 8)$ and D $(\lambda, 2,3)$ are coplanar.

$\therefore$ $[\overrightarrow{\mathrm{AB}} \,\,\,\,\overrightarrow{\mathrm{AC}} \,\,\,\, \overrightarrow{\mathrm{AD}}]=0$ $\left|\begin{array}{ccc}3 & -1 & -8 \\ -1 & \lambda-3 & -1 \\ \lambda-2 & -1 & -6\end{array}\right|=0$ $\Rightarrow[-6(\lambda-3)-1]-8(1-(\lambda-3)(\lambda-2))+(6+(\lambda$ $-2)=0$ $\Rightarrow$ $3(-6 \lambda+17)-8\left(-\lambda^2+5 \lambda-5\right)+(\lambda+4)=8$ $\Rightarrow$ $8 \lambda^2-57 \lambda+95=0$ $\therefore$ $\lambda_1 \lambda_2=\dfrac{95}{8}$

Q138

Let the plane P pass through the intersection of the planes

2x+3y-z=2

and

x+2y+3z=6

, and be perpendicular to the plane

2x+y-z+1=0

. If d is the distance of P from the point (

-

7, 1, 1), then

\mathrm{d^{2}}

is equal to :

A

\dfrac{250}{83}

B

\dfrac{250}{82}

C

\dfrac{15}{53}

D

\dfrac{25}{83}

Correct Answer

Option A

Solution

Plane $P$ , is passing through intersection of the two planes, so,

\begin{aligned} & 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\\\ & x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0 \end{aligned}

It is perpendicular with plane, $2 x+y-2+1=0$ So, $(2+\lambda) 2+(3+2 \lambda) 1+(3 \lambda-1)(-1)=0$ $\lambda=-8$ So, plane $p_1:-6 x-13 y-25 z+46=0$ Distance of plane $p$ from the point $(-7,1,1)$

\begin{aligned} & d=\frac{|+42-13-25+46|}{\sqrt{36+169+625}}=\frac{50}{\sqrt{830}} \\\\ & d^2=\frac{2500}{830}=\frac{250}{83} \end{aligned}

Q139

Let the plane

\mathrm{P}: 8 x+\alpha_{1} y+\alpha_{2} z+12=0

be parallel to the line

\mathrm{L}: \dfrac{x+2}{2}=\dfrac{y-3}{3}=\dfrac{z+4}{5}

. If the intercept of

\mathrm{P}

on the

y

-axis is 1 , then the distance between

\mathrm{P}

and

\mathrm{L}

is :

A

\dfrac{6}{\sqrt{14}}

B

\sqrt{14}

C

\sqrt{\dfrac{2}{7}}

D

\sqrt{\dfrac{7}{2}}

Correct Answer

Option B

Solution

P: $8 x+\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}+12=0$ L: $\dfrac{\mathrm{x}+2}{2}=\dfrac{\mathrm{y}-3}{3}=\dfrac{\mathrm{z}+4}{5}$ $\because \mathrm{P}$ is parallel to $\mathrm{L}$ $\Rightarrow 8(2)+\alpha_{1}(3)+5\left(\alpha_{2}\right)=0$ $\Rightarrow 3 \alpha_{1}+5\left(\alpha_{2}\right)=-16$ Also $y$ -intercept of plane $P$ is 1 $\Rightarrow \alpha_{1}=-12$ And $\alpha_{2}=4$ $\Rightarrow$ Equation of plane $\mathrm{P}$ is $2 \mathrm{x}-3 \mathrm{y}+\mathrm{z}+3=0$ $\Rightarrow$ Distance of line L from Plane $\mathrm{P}$ is $=\left|\dfrac{0-3(6)+1+3}{\sqrt{4+9+1}}\right|$ $=\sqrt{14}$

Q140

If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,

{x \over 1} = {y \over 4} = {z \over 5}

is Q, then PQ is equal to:

A

2\sqrt {42}

B

\sqrt {42}

C

6\sqrt 5

D

3\sqrt 5

Correct Answer

Option A

Solution

Equation of line PQ is

{{x - 1} \over 1} = {{y + 2} \over 4} = {{z - 3} \over 5}

Let F be ( $\lambda$ + 1, 4 $\lambda$ $-$ 2, 5 $\lambda$ + 3) Since F lies on the plane $\therefore$ 2( $\lambda$ + 1) + 3(4 $\lambda$ $-$ 2) $-$ 4(5 $\lambda$ + 3) + 22 $=$ 0 $\Rightarrow$ $-$ 6 $\lambda$ + 6 = 0 $\Rightarrow$ $\lambda$ = 1 $\therefore$ F is (2, 2, 8) PQ = 2 PF = 2

\sqrt {{1^2} + {4^2} + {5^2}}

= 2

\sqrt {42}