Practice Start Test

3D Geometry

JEE Mathematics · 279 questions · Page 11 of 28 · Click an option or "Show Solution" to reveal answer

Q101
The distance of line 3y−2z−1=0=3x−z+43y - 2z - 1 = 0 = 3x - z + 4 from the point (2, −-1, 6) is :
A 26\sqrt {26}
B 252\sqrt 5
C 262\sqrt 6
D 424\sqrt 2
Correct Answer
Option C
Solution
3y−2z−1=0=3x−z+43y - 2z - 1 = 0 = 3x - z + 4
3y−2z−1=03y - 2z - 1 = 0

D.R's ⇒\Rightarrow (0, 3, −-2)

3x−z+43x - z + 4

= 0 D.R's ⇒\Rightarrow (3, −-1, 0) Let DR's of given line are a, b, c Now, 3b −- 2c = 0 & 3a −- c = 0 ∴\therefore 6a = 3b = 2c a : b : c = 3 : 6 : 9 Any point on line 3K −- 1, 6K + 1, 9K + 1 Now, 3(3K −- 1) + 6(6K + 1)1 + 9(9K + 1) = 0 ⇒\Rightarrow K =

13{1 \over 3}

Point on line ⇒\Rightarrow (0, 3, 4) Given point (2, −-1, 6) ⇒\Rightarrow Distance =

4+16+4=26\sqrt {4 + 16 + 4} = 2\sqrt 6

Option (c)

Q102
If the mirror image of the point (2, 4, 7) in the plane 3x −- y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to :
A 54
B 50
C −-6
D −-42
Correct Answer
Option C
Solution

We know mirror image of point (x1, y1, z1) in the plane ax + by + cz = d

x−x1a=y−y1b=z−z1c=−2(ax1+by1+cz1−d)a2+b2+c2{{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = {{ - 2(a{x_1} + b{y_1} + c{z_1} - d)} \over {{a^2} + {b^2} + {c^2}}}

Here given point (2, 4, 7) and plane

3x−y+4z=23x - y + 4z = 2

then mirror image is

x−23=y−4−1=z−74=−2(6−4+28−2)9+1+16{{x - 2} \over 3} = {{y - 4} \over { - 1}} = {{z - 7} \over 4} = {{ - 2(6 - 4 + 28 - 2)} \over {9 + 1 + 16}}
⇒x−23=y−4−1=z−74=−2813\Rightarrow {{x - 2} \over 3} = {{y - 4} \over { - 1}} = {{z - 7} \over 4} = - {{28} \over {13}}

∴\therefore

x=−5813=ax = - {{58} \over {13}} = a
y=8013=by = {{80} \over {13}} = b
z=−2113=cz = - {{21} \over {13}} = c

∴\therefore

2a+b+2c2a + b + 2c
=2(−5813)+8013+2(−2113)= 2\left( { - {{58} \over {13}}} \right) + {{80} \over {13}} + 2\left( { - {{21} \over {13}}} \right)
=−116+80−4213=−7813=−6= {{ - 116 + 80 - 42} \over {13}} = {{ - 78} \over {13}} = - 6
Q103
Let x−23=y+1−2=z+3−1{{x - 2} \over 3} = {{y + 1} \over { - 2}} = {{z + 3} \over { - 1}} lie on the plane px−qy+z=5px - qy + z = 5, for some p, q ∈\in R. The shortest distance of the plane from the origin is :
A 3109\sqrt {{3 \over {109}}}
B 5142\sqrt {{5 \over {142}}}
C 571{5 \over {\sqrt {71} }}
D 1142{1 \over {\sqrt {142} }}
Correct Answer
Option B
Solution

x−23=y+1−2=z+3−1=λ\dfrac{x-2}{3}=\dfrac{y+1}{-2}=\dfrac{z+3}{-1}=\lambda (3λ+2,−2λ−1,−λ−3)(3 \lambda+2,-2 \lambda-1,-\lambda-3) lies on plane px−qy+z=5p x-q y+z=5 p(3λ+2)−q(−2λ−1)+(−λ−3)=5p(3 \lambda+2)-q(-2 \lambda-1)+(-\lambda-3)=5 λ(3p+2q−1)+(2p+q−8)=0\lambda(3 p+2 q-1)+(2 p+q-8)=0 3p+2q−1=0}p=153 p+2 q-1=0\} p=15 2p+q−8=0}q=−222 p+q-8=0\} q=-22 Equation of plane 15x+22y+z−5=015 x+22 y+z-5=0 Shortest distance from origin =∣0+0+0−5∣152+222+1=\dfrac{|0+0+0-5|}{\sqrt{15^{2}+22^{2}+1}} =5710=\dfrac{5}{\sqrt{710}} =5142=\sqrt{\dfrac{5}{142}}

Q104
Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line r→=−k^+λ(i^+j^+2k^), λ∈R\overrightarrow r = - \widehat k + \lambda \left( {\widehat i + \widehat j + 2\widehat k} \right),\,\lambda \in R. Then, which of the following points lies on T?
A (2, 1, 0)
B (1, 2, 1)
C (1, 2, 2)
D (1, 3, 2)
Correct Answer
Option B
Solution

P(1,2,1)P(1,2,1) image in plane x+2y+2z=16x+2 y+2 z=16

x−11=y−22=z−12=−2(1+2×2+2×1−16)12+22+22x−11=y−22=z−12=2Q(3,6,5)r⃗=−k^+λ(i^+j^+2k^)\begin{aligned} & \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=\frac{-2(1+2 \times 2+2 \times 1-16)}{1^{2}+2^{2}+2^{2}} \\\\ & \frac{x-1}{1}=\frac{y-2}{2}=\frac{z-1}{2}=2 \\\\ & Q(3,6,5) \\\\ & \vec{r}=-\hat{k}+\lambda(\hat{i}+\hat{j}+2 \hat{k}) \end{aligned}

AQ=3i^+6j^+6k^A Q=3 \hat{i}+6 \hat{j}+6 \hat{k}

=3(i^+2j^+2k^)n⃗=(i^+2j^+2k^)×(i^+j^+2k^)∣i^j^k^122112∣=2i^−0j^−k^\begin{aligned} & =3(\hat{i}+2 \hat{j}+2 \hat{k}) \\\\ & \vec{n}=(\hat{i}+2 \hat{j}+2 \hat{k}) \times(\hat{i}+\hat{j}+2 \hat{k}) \\\\ & \left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\1 & 2 & 2 \\1 & 1 & 2\end{array}\right| \\\\ & =2 \hat{i}-0 \hat{j}-\hat{k} \end{aligned}

Equation of plane ≡2(x−0)+0(y−0)−1(z+1)\equiv 2(x-0)+0(y-0)-1(z+1) =0=0 2x−z=12 x-z=1 Point lying on plane from the option is (1,2,1)(1,2,1) i.e., option (B)

Q105
Let the plane ax + by + cz = d pass through (2, 3, −-5) and is perpendicular to the planes 2x + y −- 5z = 10 and 3x + 5y −- 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :
A 18
B 20
C 24
D 22
Correct Answer
Option D
Solution

Equation of pane through point (2, 3, −-5) and perpendicular to planes 2x + y −- 5z = 10 and 3x + 5y −- 7z = 12 is

∣x−2y−3z+521−535−7∣=0\left| \begin{array}{lll}{x - 2} & {y - 3} & {z + 5} \\ 2 & 1 & { - 5} \\ 3 & 5 & { - 7} \end{array} \right| = 0

∴\therefore Equation of plane is

(x−2)(−7+25)−(y−3)(x - 2)( - 7 + 25) - (y - 3)
(−14+15)+(z+5) . 7=0( - 14 + 15) + (z + 5)\,.\,7 = 0

∴\therefore

18x−y+7z+2=018x - y + 7z + 2 = 0
⇒18x−y+7z=−2\Rightarrow 18x - y + 7z = - 2

∴\therefore

−18x+y−7z=2- 18x + y - 7z = 2

On comparing with

ax+by+cz=dax + by + cz = d

where d > 0 is a = −- 18, b = 1, c = −- 7, d = 2 ∴\therefore

a+7b+c+20d=22a + 7b + c + 20d = 22
Q106
If two distinct point Q, R lie on the line of intersection of the planes −x+2y−z=0 - x + 2y - z = 0 and 3x−5y+2z=03x - 5y + 2z = 0 and PQ=PR=18PQ = PR = \sqrt {18} where the point P is (1, −-2, 3), then the area of the triangle PQR is equal to :
A 2338{2 \over 3}\sqrt {38}
B 4338{4 \over 3}\sqrt {38}
C 8338{8 \over 3}\sqrt {38}
D 1523\sqrt {{{152} \over 3}}
Correct Answer
Option B
Solution

Line L is x = y = z

PQ→. (i^+j^+k^)=0\overrightarrow {PQ} .\,(\widehat i + \widehat j + \widehat k) = 0
⇒(α−3)+α+2+α−1=0\Rightarrow (\alpha - 3) + \alpha + 2 + \alpha - 1 = 0
⇒α=23\Rightarrow \alpha = {2 \over 3}

so,

T=(23,23,23)T = \left( {{2 \over 3},{2 \over 3},{2 \over 3}} \right)
PT=383PT = \sqrt {{{38} \over 3}}
⇒QT=43\Rightarrow QT = {4 \over {\sqrt 3 }}

So, Area

=(12×43×383). 2= \left( {{1 \over 2} \times {4 \over {\sqrt 3 }} \times {{\sqrt {38} } \over {\sqrt 3 }}} \right).\,2
=4383= {{4\sqrt {38} } \over 3}

sq. units

Q107
On which of the following lines lies the point of intersection of the line, x−42=y−52=z−31{{x - 4} \over 2} = {{y - 5} \over 2} = {{z - 3} \over 1} and the plane, x + y + z = 2 ?
A x−41=y−51=z−5−1{{x - 4} \over 1} = {{y - 5} \over 1} = {{z - 5} \over { - 1}}
B x−22=y−32=z+33{{x - 2} \over 2} = {{y - 3} \over 2} = {{z + 3} \over 3}
C x−11=y−32=z+4−5{{x - 1} \over 1} = {{y - 3} \over 2} = {{z + 4} \over { - 5}}
D x+33=4−y3=z+1−2{{x + 3} \over 3} = {{4 - y} \over 3} = {{z + 1} \over { - 2}}
Correct Answer
Option C
Solution

General point on the given line is x = 2λ\lambda + 4 y = 2λ\lambda + 5 z = λ\lambda + 3 Solving with plane, 2λ\lambda + 4 + 2λ\lambda + 5 + λ\lambda + 3 = 2 5λ\lambda + 12 = 2 5λ\lambda = −- 10 λ\lambda = −- 2

Q108
The acute angle between the planes P1 and P2, when P1 and P2 are the planes passing through the intersection of the planes 5x+8y+13z−29=05x + 8y + 13z - 29 = 0 and 8x−7y+z−20=08x - 7y + z - 20 = 0 and the points (2, 1, 3) and (0, 1, 2), respectively, is :
A π3{\pi \over 3}
B π4{\pi \over 4}
C π6{\pi \over 6}
D π12{\pi \over 12}
Correct Answer
Option A
Solution

Family of Plane's equation can be given by

(5+8λ)x+(8−7λ)y+(13+λ)z−(29+20λ)=0(5 + 8\lambda )x + (8 - 7\lambda )y + (13 + \lambda )z - (29 + 20\lambda ) = 0

P1 passes through (2, 1, 3)

⇒(10+16λ)+(8−7λ)+(39+3λ)−(29+20λ)=0\Rightarrow (10 + 16\lambda ) + (8 - 7\lambda ) + (39 + 3\lambda ) - (29 + 20\lambda ) = 0
⇒−8λ+28=0⇒λ=72\Rightarrow - 8\lambda + 28 = 0 \Rightarrow \lambda = {7 \over 2}

d.r, s of normal to P1

⟨33,−332,332⟩\left\langle {33,{{ - 33} \over 2},{{33} \over 2}} \right\rangle

or

⟨1,−12,12⟩\left\langle {1, - {1 \over 2},{1 \over 2}} \right\rangle

P2 passes through (0, 1, 2)

⇒8−7λ+26+2λ−(29+20λ)=0\Rightarrow 8 - 7\lambda + 26 + 2\lambda - (29 + 20\lambda ) = 0
⇒5−25λ=0\Rightarrow 5 - 25\lambda = 0
⇒λ=15\Rightarrow \lambda = {1 \over 5}

d.r, s of normal to P2

⟨335,335,665⟩\left\langle {{{33} \over 5},{{33} \over 5},{{66} \over 5}} \right\rangle

or

⟨1,1,2⟩\left\langle {1,1,2} \right\rangle

Angle between normals

=(i^−12j^+12k^). (i^+j^+2k^)32                        6= {{\left( {\widehat i - {1 \over 2}\widehat j + {1 \over 2}\widehat k} \right).\,\left( {\widehat i + \widehat j + 2\widehat k} \right)} \over {{{\sqrt 3 } \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt 6 }}
cos⁡θ=1−12+13=12\cos \theta = {{1 - {1 \over 2} + 1} \over 3} = {1 \over 2}
θ=π3\theta = {\pi \over 3}
Q109
If the lines r→=(i^−j^+k^)+λ(3j^−k^)\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right) and r→=(αi^−j^)+μ(2i^−3k^)\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right) are co-planar, then the distance of the plane containing these two lines from the point (α\alpha, 0, 0) is :
A 29{2 \over 9}
B 211{2 \over 11}
C 411{4 \over 11}
D 2
Correct Answer
Option B
Solution

∵\because Both lines are coplanar, so

∣α−10−103−120−3∣=0\left| \begin{array}{lll}{\alpha - 1} & 0 & { - 1} \\ 0 & 3 & { - 1} \\ 2 & 0 & { - 3} \end{array} \right| = 0
⇒α=53\Rightarrow \alpha = {5 \over 3}

Equation of plane containing both lines

∣x−1y+1z−103−120−3∣=0\left| \begin{array}{lll}{x - 1} & {y + 1} & {z - 1} \\ 0 & 3 & { - 1} \\ 2 & 0 & { - 3} \end{array} \right| = 0
⇒9x+2y+6z=13\Rightarrow 9x + 2y + 6z = 13

So, distance of

(53,0,0)\left( {{5 \over 3},0,0} \right)

from this plane

=281+4+36=211= {2 \over {\sqrt {81 + 4 + 36} }} = {2 \over {11}}
Q110
Let the plane P:r→ . a→=dP:\overrightarrow r \,.\,\overrightarrow a = d contain the line of intersection of two planes r→ . (i^+3j^−k^)=6\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 6 and r→ . (−6i^+5j^−k^)=7\overrightarrow r \,.\,\left( { - 6\widehat i + 5\widehat j - \widehat k} \right) = 7. If the plane P passes through the point (2,3,12)\left( {2,3,{1 \over 2}} \right), then the value of ∣13a→∣2d2{{|13\overrightarrow a {|^2}} \over {{d^2}}} is equal to :
A 90
B 93
C 95
D 97
Correct Answer
Option B
Solution
P1:x+3y−z=6{P_1}:x + 3y - z = 6
P2:−6x+5y−z=7{P_2}: - 6x + 5y - z = 7

Family of planes passing through line of intersection of P1 and P2 is given by

x(1−6λ)+y(3+5λ)+z(−1−λ)−(6+7λ)=0x(1 - 6\lambda ) + y(3 + 5\lambda ) + z( - 1 - \lambda ) - (6 + 7\lambda ) = 0

It passes through

(2,3,12)\left( {2,3,{1 \over 2}} \right)

So,

2(1−6λ)+3(3+5λ)+12(−1−λ)−(6+7λ)=02(1 - 6\lambda ) + 3(3 + 5\lambda ) + {1 \over 2}( - 1 - \lambda ) - (6 + 7\lambda ) = 0
⇒2−12λ+9+15λ−12−λ2−6−7λ=0\Rightarrow 2 - 12\lambda + 9 + 15\lambda - {1 \over 2} - {\lambda \over 2} - 6 - 7\lambda = 0
⇒92−9λ2=0⇒λ=1\Rightarrow {9 \over 2} - {{9\lambda } \over 2} = 0 \Rightarrow \lambda = 1

Required plane is

−5x+8y−2z−13=0- 5x + 8y - 2z - 13 = 0

Or

r→. (−5i^+8j^−2k^)=13\overrightarrow r .\,( - 5\widehat i + 8\widehat j - 2\widehat k) = 13
∣13a→∣2∣d∣2=132(13)2. ∣a→∣2=93{{|13\overrightarrow a {|^2}} \over {|d{|^2}}} = {{{{13}^2}} \over {{{(13)}^2}}}.\,|\overrightarrow a {|^2} = 93
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →