Practice Start Test

3D Geometry

JEE Mathematics · 279 questions · Page 16 of 28 · Click an option or "Show Solution" to reveal answer

Q151
If the lines x−11=y−22=z+31{{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} and x−a2=y+23=z−31{{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} intersect at the point P, then the distance of the point P from the plane z=az = a is :
A 28
B 22
C 10
D 16
Correct Answer
Option A
Solution
x−11=y−22=z+31=λ{{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} = \lambda

(say) &

x−a2=y+23=z−31=μ{{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} = \mu

(say) ∴\therefore

λ+1=2μ+a\lambda + 1 = 2\mu + a

...... (i)

2λ+2=3μ−22\lambda + 2 = 3\mu - 2

..... (ii)

λ−3=μ+3\lambda - 3 = \mu + 3

.... (iii) By (i) & (ii)

⇒3μ−2=4μ+2a+2\Rightarrow 3\mu - 2 = 4\mu + 2a + 2
μ=−2(1+a)\mu=-2(1+a)

&

λ=5−3a\lambda=5-3a

Put λ\lambda & μ\mu in (iii) we get

a=−9a=-9
μ=16\mu=16
λ=22\lambda=22

∴\therefore Point of intersection

≡(23,46,19)\equiv(23,46,19)

Distance from

z=−9z=-9

is 28

Q152
The shortest distance between the lines x−12=y+8−7=z−45{{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5} and x−12=y−21=z−6−3{{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}} is :
A 232\sqrt3
B 333\sqrt3
C 434\sqrt3
D 535\sqrt3
Correct Answer
Option C
Solution
r→1=i^−8j^+4k^{\overrightarrow r _1} = \widehat i - 8\widehat j + 4\widehat k
r→2=i^+2j^+6k^{\overrightarrow r _2} = \widehat i + 2\widehat j + 6\widehat k
a→=2i^−7j^+5k^\overrightarrow a = 2\widehat i - 7\widehat j + 5\widehat k
b→=2i^+j^−3k^\overrightarrow b = 2\widehat i + \widehat j - 3\widehat k

S.D.

=∣r→1−r→2a→b→∣∣a→×b→∣= {{\left| \begin{array}{lll}{{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & \begin{array}{ll}{\overrightarrow a } & {\overrightarrow b } \end{array} \end{array} \right|} \over {\left| {\overrightarrow a \times \overrightarrow b } \right|}}
[r→1−r→2a→b→]=∣0−10−22−7521−3∣\left[ \begin{array}{lll}{{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & \begin{array}{ll}{\overrightarrow a } & {\overrightarrow b } \end{array} \end{array} \right] = \left| \begin{array}{lll}0 & { - 10} & { - 2} \\ 2 & { - 7} & 5 \\ 2 & 1 & { - 3} \end{array} \right|

∴\therefore

10(−16)−2(16)=−19210( - 16) - 2(16) = - 192
∣[r→1−r→2a→b→]∣=192\left| {\left[ \begin{array}{lll}{{{\overrightarrow r }_1} - {{\overrightarrow r }_2}} & \begin{array}{ll}{\overrightarrow a } & {\overrightarrow b } \end{array} \end{array} \right]} \right| = 192
a→×b→=∣i^j^k^2−7521−3∣=16i^+16j^+16k^\overrightarrow a \times \overrightarrow b = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 2 & { - 7} & 5 \\ 2 & 1 & { - 3} \end{array} \right| = 16\widehat i + 16\widehat j + 16\widehat k
a→×b→=163\overrightarrow a \times \overrightarrow b = 16\sqrt 3

S.D.

=192163=43= {{192} \over {16\sqrt 3 }} = 4\sqrt 3
Q153
The distance of the point (−1,9,−16-1,9,-16) from the plane 2x+3y−z=52x+3y-z=5 measured parallel to the line x+43=2−y4=z−312{{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}} is :
A 132\sqrt2
B 26
C 202\sqrt2
D 31
Correct Answer
Option B
Solution

Given,

x+43=2−y4=z−312{{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}

⇒\Rightarrow

x+43=y−2−4=z−312{{x + 4} \over 3} = {{y - 2} \over -4} = {{z - 3} \over {12}}

Equation of line passing through the point P(−1,9,−16)(-1,9,-16) and parallel to line x+43=2−y4=z−312\dfrac{x+4}{3}=\dfrac{2-y}{4}=\dfrac{z-3}{12} is x+13=y−9−4=z+1612=λ\dfrac{x+1}{3}=\dfrac{y-9}{-4}=\dfrac{z+16}{12}=\lambda Any point on this line (A) =(3λ−1,−4λ+9,12λ−16)=(3 \lambda-1,-4 \lambda+9,12 \lambda-16) Point of intersection line and plane

2(3λ−1)+3(−4λ+9)−1(12λ−16)=5⇒6λ−2−12λ+27−12λ+16=5⇒λ=2∴ Point A =(5,1,8)∴ Distance (AP) =(5+1)2+(9−1)2+(−16−8)2=36+64+576=26 units \begin{aligned} & 2(3 \lambda-1)+3(-4 \lambda+9)-1(12 \lambda-16)=5 \\\\ \Rightarrow & 6 \lambda-2-12 \lambda+27-12 \lambda+16=5 \\\\ \Rightarrow & \lambda=2 \quad \\\\ \therefore & \text{ Point A }=(5,1,8) \\\\ \therefore & \text{ Distance (AP) } \\\\ = & \sqrt{(5+1)^2+(9-1)^2+(-16-8)^2}\\\\ = & \sqrt{36+64+576}\\\\ = & 26 \text{ units } \end{aligned}
Q154
The foot of perpendicular of the point (2, 0, 5) on the line x+12=y−15=z+1−1{{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}} is (α,β,γ\alpha,\beta,\gamma). Then, which of the following is NOT correct?
A αβ=−8\dfrac{\alpha}{\beta}=-8
B αβγ=415\dfrac{\alpha \beta}{\gamma}=\dfrac{4}{15}
C βγ=−5\dfrac{\beta}{\gamma}=-5
D γα=58\dfrac{\gamma}{\alpha}=\dfrac{5}{8}
Correct Answer
Option C
Solution
L:x+12=y−15=z+1−1=λ (let) \mathrm{L}: \frac{\mathrm{x}+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text{ (let) }

Let foot of perpendicular is

P(2λ−1,5λ+1,−λ−1)PA→=(3−2λ)i^−(5λ+1)j^+(6+λ)k^ Direction ratio of line ⇒b→=2i^+5j^−k^ Now, ⇒PA→⋅b→=0⇒2(3−2λ)−5(5λ+1)−(6+λ)=0⇒λ=−16P(2λ−1,5λ+1,−λ−1)≡P(α,β,γ)⇒α=2(−16)−1=−43⇒α=−43⇒β=5(−16)+1=16⇒β=16⇒γ=−λ−1=16−1⇒γ=−56\begin{aligned} & \mathrm{P}(2 \lambda-1,5 \lambda+1,-\lambda-1) \\\\ & \overrightarrow{\mathrm{PA}}=(3-2 \lambda) \hat{\mathrm{i}}-(5 \lambda+1) \hat{\mathrm{j}}+(6+\lambda) \hat{\mathrm{k}} \\\\ & \text{ Direction ratio of line } \Rightarrow \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\\\ & \text{ Now, } \Rightarrow \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{b}}=0 \\\\ & \Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0 \\\\ & \Rightarrow \lambda=\frac{-1}{6} \\\\ & \mathrm{P}(2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv \mathrm{P}(\alpha, \beta, \gamma) \\\\ & \Rightarrow \alpha=2\left(-\frac{1}{6}\right)-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3} \\\\ & \Rightarrow \beta=5\left(-\frac{1}{6}\right)+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6} \\\\ & \Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6} \end{aligned}
Q155
The shortest distance between the lines x+1=2y=−12zx+1=2y=-12z and x=y+2=6z−6x=y+2=6z-6 is :
A 3
B 52\dfrac{5}{2}
C 32\dfrac{3}{2}
D 2
Correct Answer
Option D
Solution

L1:x+11=y12=z−112L_{1}: \dfrac{x+1}{1}=\dfrac{y}{\dfrac{1}{2}}=\dfrac{z}{\dfrac{-1}{12}}

L2:x1=y+21=z−116 S.D =∣(−i^+2j^−k^)⋅(2i^−3j^+6k^)7∣=∣−2−6−67∣=2 units \begin{aligned} & L_{2}: \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}} \\\\ & \text{ S.D }=\left|\frac{(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{7}\right| \\\\ & =\left|\frac{-2-6-6}{7}\right|=2 \text{ units } \end{aligned}
Q156
The distance of the point P(4, 6, −-2) from the line passing through the point (−-3, 2, 3) and parallel to a line with direction ratios 3, 3, −-1 is equal to :
A 3
B 14\sqrt{14}
C 6\sqrt6
D 232\sqrt3
Correct Answer
Option B
Solution

AP→=7i^+4j^−5k^⇒∣AP→∣=49+16+25=90AN\overrightarrow{A P}=7 \hat{i}+4 \hat{j}-5 \hat{k} \Rightarrow|\overrightarrow{A P}|=\sqrt{49+16+25}=\sqrt{90} A N == projection of AP→\overrightarrow{A P} on b⃗=AP→⋅b⃗=21+12+519=3819\vec{b}=\overrightarrow{A P} \cdot \vec{b}=\dfrac{21+12+5}{\sqrt{19}}=\dfrac{38}{\sqrt{19}} (PN)2=(AP)2−(AN)2=90−76=14⇒PN=14(P N)^{2}=(A P)^{2}-(A N)^{2}=90-76=14 \Rightarrow P N=\sqrt{14}

Q157
Consider the lines L1L_1 and L2L_2 given by L1:x−12=y−31=z−22{L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2} L2:x−21=y−22=z−33{L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}. A line L3L_3 having direction ratios 1, −-1, −-2, intersects L1L_1 and L2L_2 at the points PP and QQ respectively. Then the length of line segment PQPQ is
A 434\sqrt3
B 262\sqrt6
C 4
D 323\sqrt2
Correct Answer
Option B
Solution

Let,

P≡(2λ+1,λ+3,2λ+2) and Q(μ+2,2μ+2, 3μ+3) d.r’s of PQ≡∴2λ−μ−11−λ−2μ−1−1=2λ−3μ−1−2∴−2λ+μ+1=λ−2μ+1 and −2λ+4μ−2=−2λ+3μ+1⇒3λ−3μ=0 and μ=3∴λ=±3 and μ=3∴P≡(7,6,8) and Q(5,8,12)∴∣PO∣=22+22+42=24=26\begin{aligned} & P \equiv(2 \lambda+1, \lambda+3,2 \lambda+2) \text{ and } Q(\mu+2,2 \mu+2 \text{, } 3 \mu+3) \\\\ & \text{ d.r's of } P Q \equiv \\\\ & \therefore \quad \frac{2 \lambda-\mu-1}{1}-\frac{\lambda-2 \mu-1}{-1}=\frac{2 \lambda-3 \mu-1}{-2} \\\\ & \therefore \quad-2 \lambda+\mu+1=\lambda-2 \mu+1 \text{ and }-2 \lambda+4 \mu-2= \\\\ & -2 \lambda+3 \mu+1 \\\\ & \Rightarrow 3 \lambda-3 \mu=0 \text{ and } \mu=3 \\\\ & \therefore \quad \lambda=\pm 3 \text{ and } \mu=3 \\\\ & \therefore \quad P \equiv(7,6,8) \text{ and } Q(5,8,12) \\\\ & \therefore|P O|=\sqrt{2^{2}+2^{2}+4^{2}}=\sqrt{24}=2 \sqrt{6} \end{aligned}
Q158
If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes x+2y+z=0x+2y+z=0 and 3y−z=33y-z=3 is (α,β,γ\alpha,\beta,\gamma), then α+β+γ\alpha+\beta+\gamma is equal to :
A 3
B 1
C −-1
D 5
Correct Answer
Option D
Solution

Direction of line

b⃗=∣i^j^k^12103−1∣=i^(−5)−j^(−1)+k^(3)=−5i^+j^+3k^\begin{aligned} \vec{b} & =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 0 & 3 & -1 \end{array}\right| \\\\ & =\hat{i}(-5)-\hat{j}(-1)+\hat{k}(3) \\\\ & =-5 \hat{i}+\hat{j}+3 \hat{k} \end{aligned}

Equation of line

x−3−5=y−21=z−13\frac{x-3}{-5}=\frac{y-2}{1}=\frac{z-1}{3}

Let foot of perpendicular be =(−5k+3,k+2,3k+1)=(-5 k+3, k+2,3 k+1) ⇒(−5k+2)(−5)+(k−7)(1)+(3k−6)(3)=0\Rightarrow(-5 k+2)(-5)+(k-7)(1)+(3 k-6)(3)=0 Or 25k−10+k−7+9k−18=025 k-10+k-7+9 k-18=0 Or k=1k=1 α+β+γ=−k+6=5\alpha+\beta+\gamma=-k+6=5

Q159
Let the plane containing the line of intersection of the planes P1 : x+(λ+4)y+z=1x+(\lambda+4)y+z=1 and P2 : 2x+y+z=22x+y+z=2 pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of the point (2λ,λ,−λ\lambda,\lambda,-\lambda) from the plane P2 is :
A 262\sqrt6
B 363\sqrt6
C 464\sqrt6
D 565\sqrt6
Correct Answer
Option B
Solution

Equation of plane passing through point of intersection of P1\mathrm{P} 1 and P2\mathrm{P} 2

P1+kP2=0(x+(λ+4)y+z−1)+k(2x+y+z−2)=0\begin{aligned} & \mathrm{P}_1+\mathrm{kP}_2 = 0 \\\\ & (\mathrm{x}+(\lambda+4) \mathrm{y}+\mathrm{z}-1)+\mathrm{k}(2 \mathrm{x}+\mathrm{y}+\mathrm{z}-2)=0 \end{aligned}

Passing through (0,1,0)(0,1,0) and (1,0,1)(1,0,1)

(λ+4−1)+k(1−2)=0(λ+3)−k=0...(1)\begin{aligned} & (\lambda+4-1)+\mathrm{k}(1-2)=0 \\\\ & (\lambda+3)-\mathrm{k}=0\quad...(1) \end{aligned}

Also passing (1,0,1)(1,0,1)

(1+1−1)+k(2+1−2)=01+k=0k=−1\begin{aligned} & (1+1-1)+\mathrm{k}(2+1-2)=0 \\\\ & 1+\mathrm{k}=0 \\\\ & \mathrm{k}=-1 \end{aligned}

put in (1)

λ+3+1=0λ=−4\begin{aligned} & \lambda+3+1=0 \\\\ & \lambda=-4 \end{aligned}

Then point (2λ,λ,−λ)(2 \lambda, \lambda,-\lambda)

(−8,−4,4)d=∣−16−4+4−26∣d=186×66=36\begin{aligned} & (-8,-4,4) \\\\ & d=\left|\frac{-16-4+4-2}{\sqrt{6}}\right| \\\\ & d=\frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=3 \sqrt{6} \end{aligned}
Q160
The distance of the point (7, −-3, −-4) from the plane passing through the points (2, −-3, 1), (−-1, 1, −-2) and (3, −-4, 2) is :
A 424\sqrt2
B 4
C 5
D 525\sqrt2
Correct Answer
Option D
Solution

A(2,−3,1),B(−1,1,−2),C(3,−4,2)A(2,-3,1), B(-1,1,-2), C(3,-4,2)

AB→=−3i^+4j^−3k^AC→=i^−j^+k^n⃗=∣i^j^k^−34−31−11∣=i^−k^\begin{aligned} & \overrightarrow{A B}=-3 \hat{i}+4 \hat{j}-3 \hat{k} \quad \overrightarrow{A C}=\hat{i}-\hat{j}+\hat{k} \\\\ & \vec{n}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\\\ -3 & 4 & -3 \\\\ 1 & -1 & 1 \end{array}\right|=\hat{i}-\hat{k} \end{aligned}

Let equation of plane is x−z+λ=0x-z+\lambda=0 passes through point A(2,−3,1)⇒λ=−1A(2,-3,1) \Rightarrow \lambda=-1 Equation of plane is x−z−1=0x-z-1=0 Distance of point (7,−3,−4)(7,-3,-4) from the plane x−z−x-z- 1=01=0 is 525 \sqrt{2}

Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →