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3D Geometry

JEE Mathematics · 279 questions · Page 13 of 28 · Click an option or "Show Solution" to reveal answer

Q121
If the shortest distance between the lines x−12=y−23=z−3λ{{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda } and x−21=y−44=z−55{{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5} is 13{1 \over {\sqrt 3 }}, then the sum of all possible value of λ\lambda is :
A 16
B 6
C 12
D 15
Correct Answer
Option A
Solution

Let

a→1=i^+2j^+3k^{\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k
a→2=2i^+4j^+5k^{\overrightarrow a _2} = 2\widehat i + 4\widehat j + 5\widehat k
p→=2i^+3j^+λk^, q→=i^+4j^+5k^\overrightarrow p = 2\widehat i + 3\widehat j + \lambda \widehat k,\,\overrightarrow q = \widehat i + 4\widehat j + 5\widehat k

∴\therefore

p→×q→=(15−4λ)i^−(10−λ)j^+5k^\overrightarrow p \times \overrightarrow q = (15 - 4\lambda )\widehat i - (10 - \lambda )\widehat j + 5\widehat k
a→2−a→1=i^+2j^+2k^{\overrightarrow a _2} - {\overrightarrow a _1} = \widehat i + 2\widehat j + 2\widehat k

∴\therefore Shortest distance

=∣(15−4λ)−2(10−λ)+10(15−4λ)2+(10−λ)2+25∣=13= \left| {{{(15 - 4\lambda ) - 2(10 - \lambda ) + 10} \over {\sqrt {{{(15 - 4\lambda )}^2} + {{(10 - \lambda )}^2} + 25} }}} \right| = {1 \over {\sqrt 3 }}
⇒3(5−2λ)2=(15−4λ)2+(10−λ)2+25\Rightarrow 3{(5 - 2\lambda )^2} = {(15 - 4\lambda )^2} + {(10 - \lambda )^2} + 25
⇒5λ2−80λ+275=0\Rightarrow 5{\lambda ^2} - 80\lambda + 275 = 0

∴\therefore Sum of values of

λ=805=16\lambda = {{80} \over 5} = 16
Q122
Let the points on the plane P be equidistant from the points (−-4, 2, 1) and (2, −-2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :
A π6{\pi \over 6}
B π4{\pi \over 4}
C π3{\pi \over 3}
D 5π12{5\pi \over 12}
Correct Answer
Option C
Solution

Let P(x, y, z) be any point on plane P1 Then

(x+4)2+(y−2)2+(z−1)2=(x−2)2+(y+2)2+(z−3)2{(x + 4)^2} + {(y - 2)^2} + {(z - 1)^2} = {(x - 2)^2} + {(y + 2)^2} + {(z - 3)^2}
⇒12x−8y+4z+4=0\Rightarrow 12x - 8y + 4z + 4 = 0
⇒3x−2y+z+1=0\Rightarrow 3x - 2y + z + 1 = 0

And

P2:2x+y+3z=0{P_2}:2x + y + 3z = 0

∴\therefore angle between P1 and P2

cos⁡θ=∣6−2+314∣⇒θ=π3\cos \theta =\left| {{{6 - 2 + 3} \over {14}}} \right| \Rightarrow \theta = {\pi \over 3}
Q123
The distance of the point (3, 2, −-1) from the plane 3x−y+4z+1=03x - y + 4z + 1 = 0 along the line 2−x2=y−32=z+11{{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1} is equal to :
A 9
B 6
C 3
D 2
Correct Answer
Option C
Solution

Line PQ is parallel to line

2−x2=y−32=z+11{{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}

∴\therefore DR of PQ = DR of line = <−-2, 2, 1> ∴\therefore Equation of line PQ passing through P(3, 2, −-1) and DR = <−-2, 2, 1> is

x−3−2=y−22=z+11{{x - 3} \over { - 2}} = {{y - 2} \over 2} = {{z + 1} \over 1}

Any General point on line PQ = (x1, y1, z1) ∴\therefore

x1−3−2=y1−22=z1+11=λ{{{x_1} - 3} \over { - 2}} = {{{y_1} - 2} \over 2} = {{{z_1} + 1} \over 1} = \lambda
⇒x1=−2λ+3\Rightarrow {x_1} = - 2\lambda + 3
y1=2λ+2{y_1} = 2\lambda + 2
z1=λ−1{z_1} = \lambda - 1

∴\therefore Point Q = (−-2λ\lambda + 3, 2λ\lambda + 2, λ\lambda −- 1) Point Q lies on the plane

3x−y+4z+1=03x - y + 4z + 1 = 0

. So point Q satisfy the equation.

3(−2λ+3)−(2λ+2)+4(λ−1)+1=03( - 2\lambda + 3) - (2\lambda + 2) + 4(\lambda - 1) + 1 = 0
⇒−6λ+9−2λ−2+4λ−4+1=0\Rightarrow - 6\lambda + 9 - 2\lambda - 2 + 4\lambda - 4 + 1 = 0
⇒−4λ+4=0\Rightarrow - 4\lambda + 4 = 0
⇒λ=1\Rightarrow \lambda = 1

∴\therefore Point Q = (−-2 ×\times 1 + 3, 2 ×\times 1 + 2, 1 −- 1) = (1, 4, 0) ∴\therefore Distance of the point P(3, 2, −-1) from the plane = Length of PQ

=(3−1)2+(2−4)2+(−1−0)2= \sqrt {{{(3 - 1)}^2} + {{(2 - 4)}^2} + {{( - 1 - 0)}^2}}
=22+(−2)2+(−1)2= \sqrt {{2^2} + {{( - 2)}^2} + {{( - 1)}^2}}
=4+4+1= \sqrt {4 + 4 + 1}
=9= \sqrt 9
=3= 3
Q124
Let QQ be the foot of perpendicular drawn from the point P(1,2,3)P(1,2,3) to the plane x+2y+z=14x+2 y+z=14. If RR is a point on the plane such that ∠PRQ=60∘\angle P R Q=60^{\circ}, then the area of △PQR\triangle P Q R is equal to :
A 32\dfrac{\sqrt{3}}{2}
B 3 \sqrt{3}
C 232 \sqrt{3}
D 3
Correct Answer
Option B
Solution
PQ=∣1+4+3−146∣=6P Q=\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}
QR=PQtan⁡60∘=63=2Q R=\frac{P Q}{\tan 60^{\circ}}=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}
 Area (△PQR)=12⋅PQ⋅QR=3\text{ Area }(\triangle P Q R)=\frac{1}{2} \cdot P Q \cdot Q R=\sqrt{3}
Q125
Let P\mathrm{P} be the plane containing the straight line x−39=y+4−1=z−7−5\dfrac{x-3}{9}=\dfrac{y+4}{-1}=\dfrac{z-7}{-5} and perpendicular to the plane containing the straight lines x2=y3=z5\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5} and x3=y7=z8\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{8}. If d\mathrm{d} is the distance of P\mathrm{P} from the point (2,−5,11)(2,-5,11), then d2\mathrm{d}^{2} is equal to :
A 1472\dfrac{147}{2}
B 96
C 323\dfrac{32}{3}
D 54
Correct Answer
Option C
Solution

Let ⟨a,b,c⟩\langle a, b, c\rangle be direction ratios of plane containing

 lines x2=y3=z5 and x3=y7=z8.∴2a+3b+5c=0… (i)  and 3a+7b+8c=0… (ii)  from eq. (i) and (ii) :a24−35=b15−16=c14−9∴ D. R’s. of plane are <11,1,−5>\begin{aligned} &\text{ lines } \frac{x}{2}=\frac{y}{3}=\frac{z}{5} \text{ and } \frac{x}{3}=\frac{y}{7}=\frac{z}{8} . \\\\ &\therefore \quad 2 a+3 b+5 c=0 \quad \ldots \text{ (i) } \\\\ &\text{ and } 3 a+7 b+8 c=0 \quad \ldots \text{ (ii) } \\\\ &\text{ from eq. (i) and (ii) }: \frac{a}{24-35}=\frac{b}{15-16}=\frac{c}{14-9} \\\\ &\therefore \text{ D. R's} \text{. of plane are }<11,1,-5> \end{aligned}

Let D.RD . R's of plane PP be

then.11a1+b1−5c1=0and9a1−b1−5c1=0Fromeq.(iii)and(iv):then. 11 a_{1}+b_{1}-5 c_{1}=0 and 9 a_{1}-b_{1}-5 c_{1}=0 From eq. (iii) and (iv) :

\begin{aligned} &\frac{a_{1}}{-5-5}=\frac{b_{1}}{-45+55}=\frac{c_{1}}{-11-9} \\\\ &\therefore \text { D.R's } \text { of plane } P \text { are } \end{aligned}

EquationplanePis:1(x−3)−1(y+4)+2(z−7)=0Equation plane P is : 1(x-3)-1(y+4)+2(z-7)=0

\Rightarrow x-y+2 z-21=0 $Distancefrompoint Distance from point (2,-5,11)is is d=\frac{|2+5+22-2|}{\sqrt{6}} \therefore d^{2}=\frac{32}{3}$

Q126
A plane EE is perpendicular to the two planes 2x−2y+z=02 x-2 y+z=0 and x−y+2z=4x-y+2 z=4, and passes through the point P(1,−1,1)P(1,-1,1). If the distance of the plane EE from the point Q(a,a,2)Q(a, a, 2) is 323 \sqrt{2}, then (PQ)2(P Q)^{2} is equal to :
A 9
B 12
C 21
D 33
Correct Answer
Option C
Solution

First plane,

P1=2x−2y+z=0{P_1} = 2x - 2y + z = 0

, normal vector

≡n‾1=(2,−2,1)\equiv {\overline n _1} = (2, - 2,1)

Second plane,

P2≡x−y+2z=4{P_2} \equiv x - y + 2z = 4

, normal vector

≡n‾2=(1,−1,2)\equiv {\overline n _2} = (1, - 1,2)

Plane perpendicular to P1 and P2 will have normal vector

n‾3{\overline n _3}

Where

n‾3=(n‾1×n‾2){\overline n _3} = \left( {{{\overline n }_1} \times {{\overline n }_2}} \right)

Hence,

n‾3=(−3,−3,0){\overline n _3} = ( - 3, - 3,0)

Equation of plane E through

P(1,−1,1)P(1, - 1,1)

and

n‾3{\overline n _3}

as normal vector

(x−1,y+1,z−1) . (−3,−3,0)=0(x - 1,y + 1,z - 1)\,.\,( - 3, - 3,0) = 0
⇒x+y=0≡E\Rightarrow x + y = 0 \equiv E

Distance of

PQ(a,a,2)PQ(a,a,2)

from

E=∣2a2∣E = \left| {{{2a} \over {\sqrt 2 }}} \right|

as given,

∣2a2∣=32⇒a= ± 3\left| {{{2a} \over {\sqrt 2 }}} \right| = 3\sqrt 2 \Rightarrow a = \, \pm \,3

Hence,

Q≡(± 3,± 3,2)Q \equiv ( \pm \,3, \pm \,3,2)

Distance

7Q=21⇒(PQ)2=217Q = \sqrt {21} \Rightarrow {(PQ)^2} = 21
Q127
The shortest distance between the lines x+7−6=y−67=z\dfrac{x+7}{-6}=\dfrac{y-6}{7}=z and 7−x2=y−2=z−6\dfrac{7-x}{2}=y-2=z-6 is :
A 2292 \sqrt{29}
B 1
C 3729\sqrt{\dfrac{37}{29}}
D 292\dfrac{\sqrt{29}}{2}
Correct Answer
Option A
Solution
L1:x+76=y−67=z−01{L_1}:{{x + 7} \over 6} = {{y - 6} \over 7} = {{z - 0} \over 1}

Any point on it

a→1(−7,6,0){\overrightarrow a _1}( - 7,6,0)

and L1 is parallel to

b→1(−6,7,1){\overrightarrow b _1}( - 6,7,1)
L2:x−7−2=y−21=z−61{L_2}:{{x - 7} \over { - 2}} = {{y - 2} \over 1} = {{z - 6} \over 1}

Any point on it

a→2(7,2,6){\overrightarrow a _2}(7,2,6)

and L2 is parallel to

b→2(−2,1,1){\overrightarrow b _2}( - 2,1,1)

Shortest distance between L1 and L2

=∣(a→2−a→1) . (b→1×b→2)∣b→1×b→2∣∣=∣(−14,4,−6) . (3,2,4)9+4+16∣= \left| {{{({{\overrightarrow a }_2} - {{\overrightarrow a }_1})\,.\,({{\overrightarrow b }_1} \times {{\overrightarrow b }_2})} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}} \right| = \left| {{{( - 14,4, - 6)\,.\,(3,2,4)} \over {\sqrt {9 + 4 + 16} }}} \right|
=229= 2\sqrt {29}

.

Q128
The length of the perpendicular from the point (1,−2,5)(1,-2,5) on the line passing through (1,2,4)(1,2,4) and parallel to the line x+y−z=0=x−2y+3z−5x+y-z=0=x-2 y+3 z-5 is :
A 212\sqrt{\dfrac{21}{2}}
B 92\sqrt{\dfrac{9}{2}}
C 732\sqrt{\dfrac{73}{2}}
D 1
Correct Answer
Option A
Solution

The line

x+y−z=0=x−2y+3z−5x + y - z = 0 = x - 2y + 3z - 5

is parallel to the vector

b→=∣i^j^k^11−11−23∣=(1,4,−3)\overrightarrow b = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 1 & 1 & { - 1} \\ 1 & { - 2} & 3 \end{array} \right| = (1,4, - 3)

Equation of the line through

P(1,2,4)P(1,2,4)

and parallel to

b→\overrightarrow b
x−11=y−2−4=z−4−3{{x - 1} \over 1} = {{y - 2} \over { - 4}} = {{z - 4} \over { - 3}}

Let

N≡(λ+1,−4λ+2,−3λ+4)N \equiv (\lambda + 1, - 4\lambda + 2, - 3\lambda + 4)
QN→=(λ,−4λ+4,−3λ−1)\overrightarrow {QN} = (\lambda , - 4\lambda + 4, - 3\lambda - 1)
QN→\overrightarrow {QN}

is perpendicular to

b→\overrightarrow {b}
⇒(λ,−4λ+4,−3λ−1) . (1,4,−3)=0\Rightarrow (\lambda , - 4\lambda + 4, - 3\lambda - 1)\,.\,(1,4, - 3) = 0
⇒λ=12\Rightarrow \lambda = {1 \over 2}

Hence

QN→=(12,2,−52)\overrightarrow {QN} = \left( {{1 \over 2},2,{{ - 5} \over 2}} \right)

and

∣QN→∣=212\left| {\overrightarrow {QN} } \right| = \sqrt {{{21} \over 2}}
Q129
A vector a⃗\vec{a} is parallel to the line of intersection of the plane determined by the vectors i^,i^+j^\hat{i}, \hat{i}+\hat{j} and the plane determined by the vectors i^−j^,i^+k^\hat{i}-\hat{j}, \hat{i}+\hat{k}. The obtuse angle between a⃗\vec{a} and the vector b⃗=i^−2j^+2k^\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k} is :
A 3π4\dfrac{3 \pi}{4}
B 2π3\dfrac{2 \pi}{3}
C 4π5\dfrac{4 \pi}{5}
D 5π6\dfrac{5 \pi}{6}
Correct Answer
Option A
Solution

If

n→1{\overrightarrow n _1}

is a vector normal to the plane determined by

i^\widehat i

and

i^+j^\widehat i + \widehat j

then

n→1=∣i^j^k^100110∣=k^{\overrightarrow n _1} = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{array} \right| = \widehat k

If

n→2{\overrightarrow n _2}

is a vector normal to the plane determined by

i^−j^\widehat i - \widehat j

and

i^+k^\widehat i + \widehat k

then

n‾2=∣i^j^k^1−10101∣=−i^−j^+k^{\overline n _2} = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 1 & { - 1} & 0 \\ 1 & 0 & 1 \end{array} \right| = - \widehat i - \widehat j + \widehat k

Vector

a→\overrightarrow a

is parallel to

n→1×n→2{\overrightarrow n _1} \times {\overrightarrow n _2}

i.e.

a→\overrightarrow a

is parallel to

∣i^j^k^001−1−11∣=i^−j^\left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 0 & 0 & 1 \\ { - 1} & { - 1} & 1 \end{array} \right| = \widehat i - \widehat j

Given

b→=i^−2j^+2k^\overrightarrow b = \widehat i - 2\widehat j + 2\widehat k

Cosine of acute angle between

a→\overrightarrow a

and

b→=∣a→ . b→∣a→∣ . ∣b→∣∣=12\overrightarrow b = \left| {{{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow a |\,.\,|\overrightarrow b |}}} \right| = {1 \over {\sqrt 2 }}

Obtuse angle between

a→\overrightarrow a

and

b→=3π4\overrightarrow b = {{3\pi } \over 4}
Q130
If the plane PP passes through the intersection of two mutually perpendicular planes 2x+ky−5z=12 x+k y-5 z=1 and $$3 k x-k y+z=5, k
A 111\dfrac{1}{11}
B 511\dfrac{5}{11}
C 6
D 7
Correct Answer
Option D
Solution
P1:2x+ky−5z=1{P_1}:2x + ky - 5z = 1
P2:3kx−ky+z=5{P_2}:3kx - ky + z = 5

∵\because

P1 ⊥ P2⇒6k−k2+5=0{P_1}\, \bot \,{P_2} \Rightarrow 6k - {k^2} + 5 = 0
⇒k=1,5\Rightarrow k = 1,5

∵\because

kk

\therefore

k = 1

{P_1}:2x + y - 5z = 1

{P_2}:3x - y + z = 5

P:(2x + y - 5z - 1) + \lambda (3x - y + z - 5) = 0

Positivex−axisintercept=1Positive x-axis intercept = 1

\Rightarrow {{1 + 5\lambda } \over {2 + 3\lambda }} = 1

\Rightarrow \lambda = {1 \over 2}

\therefore

P:7x + y - 4z = 7$$ y intercept = 7.

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