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3D Geometry

JEE Mathematics · 279 questions · Page 21 of 28 · Click an option or "Show Solution" to reveal answer

Q201
Let (α,β,γ)(\alpha, \beta, \gamma) be the mirror image of the point (2,3,5)(2,3,5) in the line x−12=y−23=z−34\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}. Then, 2α+3β+4γ2 \alpha+3 \beta+4 \gamma is equal to
A 32
B 31
C 33
D 34
Correct Answer
Option C
Solution
∵PR→⊥(2,3,4)∴PR→⋅(2,3,4)=0(α−2,β−3,γ−5)⋅(2,3,4)=0⇒2α+3β+4γ=4+9+20=33\begin{aligned} & \because \overrightarrow{\mathrm{PR}} \perp(2,3,4) \\ & \therefore \overrightarrow{\mathrm{PR}} \cdot(2,3,4)=0 \\ & (\alpha-2, \beta-3, \gamma-5) \cdot(2,3,4)=0 \\ & \Rightarrow 2 \alpha+3 \beta+4 \gamma=4+9+20=33 \end{aligned}
Q202
The shortest distance, between lines L1L_1 and L2L_2, where L1:x−12=y+1−3=z+42L_1: \dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+4}{2} and L2L_2 is the line, passing through the points A(−4,4,3),B(−1,6,3)\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3) and perpendicular to the line x−3−2=y3=z−11\dfrac{x-3}{-2}=\dfrac{y}{3}=\dfrac{z-1}{1}, is
A 141221\dfrac{141}{\sqrt{221}}
B 24117\dfrac{24}{\sqrt{117}}
C 42117\dfrac{42}{\sqrt{117}}
D 121221\dfrac{121}{\sqrt{221}}
Correct Answer
Option A
Solution
L2=x+43=y−42=z−30∴S.D=∣x2−x1y2−y1z2−z12−32320∣∣n1→×n2→∣=∣5−5−72−32320∣∣n1→×n2→∣=141∣−4i^+6j^+13k^∣=14116+36+169=141221\begin{aligned} & \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \\ & \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc} \mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|} \\ & =\frac{\left|\begin{array}{ccc} 5 & -5 & -7 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|} \\ & =\frac{141}{|-4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}|} \\ & =\frac{141}{\sqrt{16+36+169}} \\ & =\frac{141}{\sqrt{221}} \end{aligned}
Q203
Let OO be the origin and the position vectors of AA and BB be 2i^+2j^+k^2 \hat{i}+2 \hat{j}+\hat{k} and 2i^+4j^+4k^2 \hat{i}+4 \hat{j}+4 \hat{k} respectively. If the internal bisector of ∠AOB\angle \mathrm{AOB} meets the line AB\mathrm{AB} at C\mathrm{C}, then the length of OCO C is
A 3234\dfrac{3}{2} \sqrt{34}
B 2331\dfrac{2}{3} \sqrt{31}
C 2334\dfrac{2}{3} \sqrt{34}
D 3231\dfrac{3}{2} \sqrt{31}
Correct Answer
Option C
Solution
 length of OC=1363=2343\text{ length of } O C=\frac{\sqrt{136}}{3}=\frac{2 \sqrt{34}}{3}
Q204
Let the line L\mathrm{L} intersect the lines x−2=−y=z−1,2(x+1)=2(y−1)=z+1x-2=-y=z-1,2(x+1)=2(y-1)=z+1 and be parallel to the line x−23=y−11=z−22\dfrac{x-2}{3}=\dfrac{y-1}{1}=\dfrac{z-2}{2}. Then which of the following points lies on L\mathrm{L} ?
A (−13,1,−1)\left(-\dfrac{1}{3}, 1,-1\right)
B (−13,−1,1)\left(-\dfrac{1}{3},-1,1\right)
C (−13,−1,−1)\left(-\dfrac{1}{3},-1,-1\right)
D (−13,1,1)\left(-\dfrac{1}{3}, 1,1\right)
Correct Answer
Option A
Solution
L1:x−21=y−1=z−11=λL2:x+1(1/2)=y−1(1/2)=z+11=μ\begin{aligned} & L_1: \frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{1}=\lambda \\ & L_2: \frac{x+1}{(1 / 2)}=\frac{y-1}{(1 / 2)}=\frac{z+1}{1}=\mu \end{aligned}

Any point of

L1L_1

and

L2L_2

will be

(λ+2,−λ,λ+1)(\lambda+2,-\lambda, \lambda+1)

and

(μ2−1,μ2+1,μ−1)\left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)

Now Dr of line

$ Now

\frac{\lambda-\frac{\mu}{3}+3}{3}=\frac{-\lambda-\frac{\mu}{2}-1}{1}=\frac{\lambda-\mu+2}{2}

\left. {\matrix{ {\lambda - {\mu \over 3} + 3 = - 3\lambda - {{3\mu } \over 2} - 3\,\,\,...(1)} \cr {2\left( {\lambda - {\mu \over 3} + 3} \right) = 3(\lambda - \mu + 2)\,\,...(2)} \cr } } \right\}\lambda = {{ - 4} \over 3},\mu = {{ - 2} \over 3}

\therefore \quad

PointswillbePoints will be

\left(\frac{2}{3}, \frac{4}{3}, \frac{-1}{3}\right)

andand

\left(\frac{-4}{3}, \frac{2}{3}, \frac{-5}{3}\right)

\therefore \quad L

willbewill be

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

\therefore \quad\left(\frac{-1}{3}, 1,-1\right)

willsatisfywill satisfy

L$$

Q205
Let the point, on the line passing through the points P(1,−2,3)P(1,-2,3) and Q(5,−4,7)Q(5,-4,7), farther from the origin and at a distance of 9 units from the point PP, be (α,β,γ)(\alpha, \beta, \gamma). Then α2+β2+γ2\alpha^2+\beta^2+\gamma^2 is equal to :
A 150
B 155
C 160
D 165
Correct Answer
Option B
Solution

Line through

PQP Q
x−14=y+2−2=z−34\frac{x-1}{4}=\frac{y+2}{-2}=\frac{z-3}{4}

Any point on

PQP Q

. be

R(4λ+1,−2λ−2,4λ+3)R(4 \lambda+1,-2 \lambda-2,4 \lambda+3)
PR=9P R=9

unit

(PR)2=81(P R)^2=81
(4λ+1−1)2+(−2λ−2+2)2+(4λ+3−3)2=81(4 \lambda+1-1)^2+(-2 \lambda-2+2)^2+(4 \lambda+3-3)^2=81
16λ2+4λ2+16λ2=8116 \lambda^2+4 \lambda^2+16 \lambda^2=81
36λ2=8136 \lambda^2=81
λ=±96=±32\lambda= \pm \frac{9}{6}= \pm \frac{3}{2}
∴R\therefore R

can be

(7,−5,9)(7,-5,9)

or

(−5,1,−3)(-5,1,-3)

Distance from origin for both points be

49+25+81\sqrt{49+25+81}

and

25+1+9=35\sqrt{25+1+9}=\sqrt{35}

∴\therefore Distance of

(7,−5,9)(7,-5,9)

is farthest from origin

∴(α,β,γ)=(7,−5,9)\therefore(\alpha, \beta, \gamma)=(7,-5,9)

Now

72+(−5)2+92=1557^2+(-5)^2+9^2=155
Q206
Let P\mathrm{P} be the point of intersection of the lines x−21=y−45=z−21\dfrac{x-2}{1}=\dfrac{y-4}{5}=\dfrac{z-2}{1} and x−32=y−23=z−32\dfrac{x-3}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{2}. Then, the shortest distance of P\mathrm{P} from the line 4x=2y=z4 x=2 y=z is
A 3147\dfrac{3 \sqrt{14}}{7}
B 5147\dfrac{5 \sqrt{14}}{7}
C 147\dfrac{\sqrt{14}}{7}
D 6147\dfrac{6 \sqrt{14}}{7}
Correct Answer
Option A
Solution
L1:x−21=y−45=z−21L2:x−32=y−23=z−32\begin{aligned} & L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1} \\ & L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2} \end{aligned}

Point of intersection of

L1L_1

and

L2L_2

is

(−1,1,−1)(-1,1,-1)

Distance of point

PP

from

L3:4x=2y=zL_3: 4 x=2 y=z
L3:x14=y12=z1L_3: \frac{x}{\frac{1}{4}}=\frac{y}{\frac{1}{2}}=\frac{z}{1}

Any point on

L3L_3

be

(λ4,λ2,λ)PR:⟨λ4+1,λ2−1,λ+1⟩∵PR⊥⟨14,12,1⟩\begin{aligned} & \left(\frac{\lambda}{4}, \frac{\lambda}{2}, \lambda\right) \\ & P R:\left\langle\frac{\lambda}{4}+1, \frac{\lambda}{2}-1, \lambda+1\right\rangle \\ & \because P R \perp\left\langle\frac{1}{4}, \frac{1}{2}, 1\right\rangle \end{aligned}
⇒(λ4+1)14+12(λ2−1)+λ+1=0λ16+14+λ4−12+λ+1=0⇒λ=−47∴R(−17,−27,−47) Now RP:(−17+1)2+(−27−1)2+(−47+1)2=3649+8149+949=1267=3147\begin{aligned} & \Rightarrow\left(\frac{\lambda}{4}+1\right) \frac{1}{4}+\frac{1}{2}\left(\frac{\lambda}{2}-1\right)+\lambda+1=0 \\ & \quad \frac{\lambda}{16}+\frac{1}{4}+\frac{\lambda}{4}-\frac{1}{2}+\lambda+1=0 \\ & \Rightarrow \quad \lambda=\frac{-4}{7} \\ & \therefore \quad R\left(\frac{-1}{7}, \frac{-2}{7}, \frac{-4}{7}\right) \\ & \text{ Now } R P: \sqrt{\left(\frac{-1}{7}+1\right)^2+\left(\frac{-2}{7}-1\right)^2+\left(\frac{-4}{7}+1\right)^2} \\ & \quad=\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}=\frac{3 \sqrt{14}}{7} \end{aligned}
Q207
If the shortest distance between the lines x−λ2=y−43=z−34\dfrac{x-\lambda}{2}=\dfrac{y-4}{3}=\dfrac{z-3}{4} and x−24=y−46=z−78\dfrac{x-2}{4}=\dfrac{y-4}{6}=\dfrac{z-7}{8} is 1329\dfrac{13}{\sqrt{29}}, then a value of λ\lambda is :
A 1325\dfrac{13}{25}
B 1
C −-1
D −1325-\dfrac{13}{25}
Correct Answer
Option B
Solution
l⃗1=2i^+3j^+4k^l⃗2=4i^+6j^+8k^\begin{aligned} & \vec{l}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\ & \vec{l}_2=4 \hat{i}+6 \hat{j}+8 \hat{k} \end{aligned}
S.D.=∣(2i^+3j^+4k^)×((λ−2)i^−4k^)∣∣2i^+3j^+4k^∣S . D .=\frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((\lambda-2) \hat{i}-4 \hat{k})|}{|2 \hat{i}+3 \hat{j}+4 \hat{k}|}
∣i^j^k^234λ−20−4∣=−12i^−j^(−8−4λ+8)+k^(6−3λ)=−12i^+4λj^+(6−3λ)k^144+16λ2+(6−3λ)229=1329144+16λ2+(6−3λ)2=169⇒16λ2+9λ2+36−36λ+144−169=0\begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ \lambda-2 & 0 & -4 \end{array}\right|=-12 \hat{i}-\hat{j}(-8-4 \lambda+8)+\hat{k}(6-3 \lambda) \\ & =-12 \hat{i}+4 \lambda \hat{j}+(6-3 \lambda) \hat{k} \\ & \frac{\sqrt{144+16 \lambda^2+(6-3 \lambda)^2}}{\sqrt{29}}=\frac{13}{\sqrt{29}} \\ & 144+16 \lambda^2+(6-3 \lambda)^2=169 \\ & \Rightarrow 16 \lambda^2+9 \lambda^2+36-36 \lambda+144-169=0 \\ \end{aligned}
⇒25λ2−36λ+11=0⇒25λ2−25λ−11λ+11=0⇒(25λ−11)(λ−1)=0⇒λ=1,1125\begin{aligned} & \Rightarrow 25 \lambda^2-36 \lambda+11=0 \\ & \Rightarrow 25 \lambda^2-25 \lambda-11 \lambda+11=0 \\ & \Rightarrow(25 \lambda-11)(\lambda-1)=0 \\ & \Rightarrow \lambda=1, \frac{11}{25} \end{aligned}
Q208
Let P(x,y,z)P(x, y, z) be a point in the first octant, whose projection in the xyx y-plane is the point QQ. Let OP=γO P=\gamma; the angle between OQO Q and the positive xx-axis be θ\theta; and the angle between OPO P and the positive zz-axis be ϕ\phi, where OO is the origin. Then the distance of PP from the xx-axis is
A γ1−sin⁡2ϕcos⁡2θ\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}
B γ1+cos⁡2θsin⁡2ϕ\gamma \sqrt{1+\cos ^2 \theta \sin ^2 \phi}
C γ1+cos⁡2ϕsin⁡2θ\gamma \sqrt{1+\cos ^2 \phi \sin ^2 \theta}
D γ1−sin⁡2θcos⁡2ϕ\gamma \sqrt{1-\sin ^2 \theta \cos ^2 \phi}
Correct Answer
Option A
Solution
OP→=xi^+yj^+zk^OQ→=xi^+yj^∣OP∣=γ=x2+y2+z2cos⁡θ=xx2+y2⇒cos⁡2θ=x2γ2−z2=x2γ2−γ2cos⁡2ϕcos⁡ϕ=zx2+y2+z2=zγ Distance of P from x-axis =y2+z2d=γ2−x2⇒x2=γ2sin⁡2ϕcos⁡2θ⇒d=γ2−γ2sin⁡2ϕcos⁡2θ=γ1−sin⁡2ϕcos⁡2θ\begin{aligned} & \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\ & \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\ & |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\ & \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\ & \cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\gamma} \\ & \text{ Distance of } P \text{ from } x \text{-axis }=\sqrt{y^2+z^2} \\ & d=\sqrt{\gamma^2-x^2} \\ & \Rightarrow x^2=\gamma^2 \sin ^2 \phi \cos ^2 \theta \\ & \Rightarrow d=\sqrt{\gamma^2-\gamma^2 \sin ^2 \phi \cos ^2 \theta} \\ & =\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta} \end{aligned}
Q209
The number of distinct real values of λ\lambda for which the lines x−11=y−22=z+3λ2{{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over {{\lambda ^2}}} and x−31=y−2λ2=z−12{{x - 3} \over 1} = {{y - 2} \over {{\lambda ^2}}} = {{z - 1} \over 2} are coplanar is :
A 4
B 1
C 2
D 3
Correct Answer
Option D
Solution

As lines are coplanar, so

∣3−12−21−(−3)12λ21λ22∣\left| \begin{array}{lll}{3 - 1} & {2 - 2} & {1 - \left( { - 3} \right)} \\ 1 & 2 & {{\lambda ^2}} \\ 1 & {{\lambda ^2}} & 2 \end{array} \right|

= 0 ⇒\Rightarrow

∣20412λ21λ22∣\left| \begin{array}{lll}2 & 0 & 4 \\ 1 & 2 & {{\lambda ^2}} \\ 1 & {{\lambda ^2}} & 2 \end{array} \right|

= 0 ⇒\Rightarrow 2(4 −- λ\lambda4) + 4(λ\lambda2 −- 2) = 0 ⇒\Rightarrow 4 −- λ\lambda4 + 2λ\lambda2 −- 4 = 0 ⇒\Rightarrow λ\lambda2(λ\lambda2 −- 2) = 0 ⇒\Rightarrow λ\lambda = 0,

2,−2\sqrt 2 , - \sqrt 2

∴\therefore 3 distinct real values are possible.

Q210
If the shortest distance between the lines L1:r⃗=(2+λ)i^+(1−3λ)j^+(3+4λ)k^,λ∈RL2:r⃗=2(1+μ)i^+3(1+μ)j^+(5+μ)k^,μ∈R\begin{array}{ll} L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\ L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, & \mu \in \mathbb{R} \end{array} is mn\dfrac{m}{\sqrt{n}}, where gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1, then the value of m+nm+n equals
A 384
B 387
C 390
D 377
Correct Answer
Option B
Solution
L1:r⃗=(2+λ)i^+(1−3λ)j^+(3+4λ)k^L1=2i^+j^+3k^+λ(i^−3j^+4k^)L2:r⃗=2i^+3j^+5k^+μ(2i^+3j^+k^)a⃗1=2i^+j^+3k^a⃗2=2i^+3j^+5k^a⃗2−a⃗1=2j^+2k^b⃗1=i^−3j^+4k^,b⃗2=2i^+3j^+k^b⃗1×b⃗2=∣i^j^k^1−34231∣\begin{aligned} & L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \\ & L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \\ & L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \\ & \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \\ & \vec{a}_2=2 \hat{i}+3 \hat{j}+5 \hat{k} \\ & \vec{a}_2-\vec{a}_1=2 \hat{j}+2 \hat{k} \\ & \vec{b}_1=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}_2=2 \hat{i}+3 \hat{j}+\hat{k} \\ & \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{array}\right| \end{aligned}
i^(−3−12)−j^(1−8)+k^(3+6)=−15i^+7j^+9k^∣b⃗1×b⃗2∣=225+49+81∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣=14+18355=32355m+n=387\begin{aligned} & \hat{i}(-3-12)-\hat{j}(1-8)+\hat{k}(3+6) \\ & =-15 \hat{i}+7 \hat{j}+9 \hat{k} \\ & \left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{225+49+81} \\ & \left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\frac{14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\ & m+n=387 \end{aligned}
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