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3D Geometry

JEE Mathematics · 279 questions · Page 9 of 28 · Click an option or "Show Solution" to reveal answer

Q81
If for a > 0, the feet of perpendiculars from the points A(a, −-2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, −-a, −-1) and D respectively, then the length of line segment CD is equal to :
A 41\sqrt {41}
B 55\sqrt {55}
C 31\sqrt {31}
D 66\sqrt {66}
Correct Answer
Option D
Solution

Let ϕ\phi is the angle between

AB→\overrightarrow {AB}

and

n→\overrightarrow n

. CD = AR = | AB |sinϕ\phi CD = | AB |

1−cos⁡2ϕ\sqrt {1 - {{\cos }^2}\phi }

CD = | AB |

1−(AB→. n→∣AB∣)2\sqrt {1 - {{\left( {{{\overrightarrow {AB} .\,\overrightarrow n } \over {|AB|}}} \right)}^2}}
=(AB)2−(AB→ . n→)2= \sqrt {{{(AB)}^2} - {{(\overrightarrow {AB} \,.\,\overrightarrow n )}^2}}

[

cos⁡ϕ=AB→ . n→∣n→∣∣AB→∣\cos \phi = {{\overrightarrow {AB} \,.\,\overrightarrow n } \over {|\overrightarrow n ||\overrightarrow {AB} |}}

]

∣AB→∣ =ai^−(2a+4)j^−2k^|\overrightarrow {AB} |\, = a\widehat i - (2a + 4)\widehat j - 2\widehat k
AB→ . n→=la−(2a+4)−2n\overrightarrow {AB} \,.\,\overrightarrow n = la - (2a + 4) - 2n

C on plane (0)l −- am −- n = 0 ..... (1) Also,

AC→\overrightarrow {AC}

||

n→\overrightarrow n
al=−am=4n{a \over l} = {{ - a} \over m} = {4 \over n}

m = −-l & an + 4m = 0 ..... (2) From (1) and (2) a2m + an = 0

4m+an=0‾\underline {4m + an = 0}

(a2 −- 4)m = 0 ⇒\Rightarrow a = 2 2m + n = 0 .... (1) m + l = 0 l2 + m2 + n2 = 1 m2 + m2 + 4m2 = 1 m2 =

16{1 \over 6}

m =

16{1 \over {\sqrt 6 }}

n =

−26{{ - 2} \over {\sqrt 6 }}

l =

−16{{ - 1} \over {\sqrt 6 }}

Now,

AB→ . n→=2(−16)−8(−16)−2(−26)\overrightarrow {AB} \,.\,\overrightarrow n = 2\left( {{{ - 1} \over {\sqrt 6 }}} \right) - 8\left( {{{ - 1} \over {\sqrt 6 }}} \right) - 2\left( {{{ - 2} \over {\sqrt 6 }}} \right)
=−2−8+46=−6= {{ - 2 - 8 + 4} \over {\sqrt 6 }} = - \sqrt 6
∣AB→∣ =4+64+4=72|\overrightarrow {AB} |\, = \sqrt {4 + 64 + 4} = \sqrt {72}
CD=72−6CD = \sqrt {72 - 6}
CD=66CD = \sqrt {66}
Q82
If the foot of the perpendicular from point (4, 3, 8) on the line L1:x−al=y−23=z−b4{L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}, l ≠\ne 0 is (3, 5, 7), then the shortest distance between the line L1 and line L2:x−23=y−44=z−55{L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5} is equal to :
A 16{1 \over {\sqrt 6 }}
B 12{1 \over 2}
C 13{1 \over {\sqrt 3 }}
D 23\sqrt {{2 \over 3}}
Correct Answer
Option A
Solution

(3, 5, 7) lie on given line L1

3−al=33=7−b4{{3 - a} \over l} = {3 \over 3} = {{7 - b} \over 4}
7−b4=1⇒b=3{{7 - b} \over 4} = 1 \Rightarrow b = 3
3−al=1⇒3−a=l{{3 - a} \over l} = 1 \Rightarrow 3 - a = l

M(4, 3, 8) N(3, 5, 7) DR'S of MN = (1, −-2, 1) MN ⊥\bot line L1 (1)(l) + (−-2)(3) + 4(1) = 0 ⇒\Rightarrow l = 2 a = 1 a = 1, b = 3, l = 2

x−12=y−23=z−34{{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over 4}
x−23=y−44=z−55{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}
A= <1,2,3>A = \, < 1,2,3 >
B= <2,4,5>B = \, < 2,4,5 >
AB→= <1,2,2>\overrightarrow {AB} = \, < 1,2,2 >
p→=2i^+3j^+4k^\overrightarrow p = 2\widehat i + 3\widehat j + 4\widehat k
q→=3i^+4j^+5k^\overrightarrow q = 3\widehat i + 4\widehat j + 5\widehat k
p→ ×q→=−i^+2j^−k^\overrightarrow p \, \times \overrightarrow q = - \widehat i + 2\widehat j - \widehat k

Shortest distance =

∣AB→ . (p→×q→)∣p→×q→∣∣ =16\left| {{{\overrightarrow {AB} \,.\,(\overrightarrow p \times \overrightarrow q )} \over {|\overrightarrow p \times \overrightarrow q |}}} \right|\, = {1 \over {\sqrt 6 }}
Q83
If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0), (0, 42, 0) and (0, 0, 42), then the value of the expression 3+x−11(y−19)2(z−12)2+y−19(x−11)2(z−12)2+z−12(x−11)2(y−19)2−x+y+z14(x−11)(y−19)(z−12)3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {{x + y + z} \over {14(x - 11)(y - 19)(z - 12)}} is equal to :
A 3
B 39
C −-45
D 0
Correct Answer
Option A
Solution

From intercept from, equation of plane is x + y + z = 42 ⇒\Rightarrow (x −- 11) + (y −- 19) + (z −- 12) = 0 let a = x −- 11, b = y −- 19, c = z −- 12 a + b + c = 0 Now, given expression is

3+ab2c2+ba2c2+ca2b2−4214abc3 + {a \over {{b^2}{c^2}}} + {b \over {{a^2}{c^2}}} + {c \over {{a^2}{b^2}}} - {{42} \over {14abc}}
3+a3+b3+c3−3abca2b2c23 + {{{a^3} + {b^3} + {c^3} - 3abc} \over {{a^2}{b^2}{c^2}}}

If a + b + c = 0 ⇒\Rightarrow a3 + b3 + c3 = 3 abc ∴\therefore

3+0a2b2c23 + {0 \over {{a^2}{b^2}{c^2}}}
=3= 3
Q84
The equation of the plane which contains the y-axis and passes through the point (1, 2, 3) is :
A x + 3z = 0
B 3x −- z = 0
C x + 3z = 10
D 3x + z = 6
Correct Answer
Option B
Solution

Let the equation of the plane is a (x −- 1) + b(y −- 2) + c(z −- 3) = 0 Y-axis lies on it.

D.R.'s of y-axis are 0, 1, 0 ∴\therefore 0.a + 1.b + 0.c = 0 ⇒\Rightarrow b = 0 ∴\therefore Equation of plane is a(x −- 1) + c(z −- 3) = 0 x = 0, z = 0 also satisfy it −-a −-3c = 0 ⇒\Rightarrow a = −-3c −-3c (x −- 1) + c (z −- 3) = 0 −-3 + 3 + z −- 3 = 0 3x −- z = 0

Q85
If the equation of plane passing through the mirror image of a point (2, 3, 1) with respect to line x+12=y−31=z+2−1{{x + 1} \over 2} = {{y - 3} \over 1} = {{z + 2} \over { - 1}} and containing the line x−23=1−y2=z+11{{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1} is α\alphax + β\betay + γ\gammaz = 24, then α\alpha + β\beta + γ\gamma is equal to :
A 21
B 19
C 18
D 20
Correct Answer
Option B
Solution

Let point M is (2λ\lambda −- 1, λ\lambda + 3, −- λ\lambda −- 2) D.R.'s of AM line are < 2λ\lambda −- 1 −- 2, λ\lambda + 3 −- 3, −-λ\lambda −- 2 −- 1> = < 2λ\lambda −- 3, λ\lambda, −-λ\lambda −-3 > AM ⊥\bot line L1 ∴\therefore

2(2λ−3)+1(λ)−1(−λ−3)=02(2\lambda - 3) + 1(\lambda ) - 1( - \lambda - 3) = 0
6λ=3,λ=126\lambda = 3,\lambda = {1 \over 2}

∴\therefore

M≡(0,72,−52)M \equiv \left( {0,{7 \over 2},{{ - 5} \over 2}} \right)

M is mid-point of A & B

M=A+B2M = {{A + B} \over 2}

B = 2M −- A B ≡\equiv (−-2, 4, −-6) Now we have to find equation of plane passing through B(−-2, 4, −-6) & also containing the line

x−23=1−y2=z+11{{x - 2} \over 3} = {{1 - y} \over 2} = {{z + 1} \over 1}

....... (1)

x−23=y−1−2=z+11{{x - 2} \over 3} = {{y - 1} \over { - 2}} = {{z + 1} \over 1}

Point P on line is (2, 1, −-1)

b→2{\overrightarrow b _2}

of line L2 is 3, −-2, 1

n→∣∣(b→2×PB→)\overrightarrow n ||({\overrightarrow b _2} \times \overrightarrow {PB} )
b→2=3i^−2j^+k^{\overrightarrow b _2} = 3\widehat i - 2\widehat j + \widehat k
PB→=−4i^+3j^−5k^\overrightarrow {PB} = - 4\widehat i + 3\widehat j - 5\widehat k
n→=7i^+11j^+k^\overrightarrow n = 7\widehat i + 11\widehat j + \widehat k

∴\therefore equation of plane is

r→ . n→=a→ . n→\overrightarrow r \,.\,\overrightarrow n = \overrightarrow a \,.\,\overrightarrow n
r→ . (7i^+11j^+k^)=(−2i^+4j^−6k^).(7i^+11j^+k^)\overrightarrow r \,.\,(7\widehat i + 11\widehat j + \widehat k) = ( - 2\widehat i + 4\widehat j - 6\widehat k).(7\widehat i + 11\widehat j + \widehat k)

7x + 11y + z = −-14 + 44 −-6 7x + 11y + z = 24 ∴\therefore α\alpha = 7 β\beta = 11 γ\gamma = 1 ∴\therefore α\alpha + β\beta + γ\gamma = 19

Q86
The lines x = ay −- 1 = z −- 2 and x = 3y −- 2 = bz −- 2, (ab ≠\ne 0) are coplanar, if :
A b = 1, a∈\inR −- {0}
B a = 1, b∈\inR −- {0}
C a = 2, b = 2
D a = 2, b = 3
Correct Answer
Option A
Solution

Lines are

x=ay−1=z−2x = ay - 1 = z - 2

∴\therefore

x1=y−1a1a=z−21{x \over 1} = {{y - {1 \over a}} \over {{1 \over a}}} = {{z - 2} \over 1}

.... (i) and

x=3y−2=bz−2x = 3y - 2 = bz - 2

∴\therefore

x1=y−2313=z−2b1b{x \over 1} = {{y - {2 \over 3}} \over {{1 \over 3}}} = {{z - {2 \over b}} \over {{1 \over b}}}

.... (ii) ∴\therefore lines are co-planar ∴\therefore

∣0−1a+23−2+2b11a11131b∣=0\left| \begin{array}{lll}0 & { - {1 \over a} + {2 \over 3}} & { - 2 + {2 \over b}} \\ 1 & {{1 \over a}} & 1 \\ 1 & {{1 \over 3}} & {{1 \over b}} \end{array} \right| = 0

∴\therefore

∣023−1a2b−201a−131−1b1131b∣=0\left| \begin{array}{lll}0 & {{2 \over 3} - {1 \over a}} & {{2 \over b} - 2} \\ 0 & {{1 \over a} - {1 \over 3}} & {1 - {1 \over b}} \\ 1 & {{1 \over 3}} & {{1 \over b}} \end{array} \right| = 0

∴\therefore

1a−1ab=0{1 \over a} - {1 \over {ab}} = 0

⇒\Rightarrow b = 1 and a

∈\in

R −- {0}

Q87
Let L be the line of intersection of planes r→.(i^−j^+2k^)=2\overrightarrow r .(\widehat i - \widehat j + 2\widehat k) = 2 and r→.(2i^+j^−k^)=2\overrightarrow r .(2\widehat i + \widehat j - \widehat k) = 2. If P(α,β,γ)P(\alpha ,\beta ,\gamma ) is the foot of perpendicular on L from the point (1, 2, 0), then the value of 35(α+β+γ)35(\alpha + \beta + \gamma ) is equal to :
A 101
B 119
C 143
D 134
Correct Answer
Option B
Solution
P1:x−y+2z=2{P_1}:x - y + 2z = 2
P2:2x+y−3=2{P_2}:2x + y - 3 = 2

Let line of Intersection of planes P1 and P2 cuts xy plane in point Q. ⇒\Rightarrow z-coordinate of point Q is zero

⇒x−y=2and 2x+y=2}⇒x=43,y=−23\Rightarrow \left. \begin{array}{ll}{x - y = 2} \\ {and\,2x + y = 2} \end{array} \right\} \Rightarrow x = {4 \over 3},y = {{ - 2} \over 3}
⇒Q(43,−23,0)\Rightarrow Q\left( {{4 \over 3},{{ - 2} \over 3},0} \right)

Vector parallel to the line of intersection

a→=∣i^j^k^1−1221−1∣=−i^+5j^+3k^\overrightarrow a = \left| \begin{array}{lll}{\widehat i} & {\widehat j} & {\widehat k} \\ 1 & { - 1} & 2 \\ 2 & 1 & { - 1} \end{array} \right| = - \widehat i + 5\widehat j + 3\widehat k

Equation of Line of intersection

x−43−1=y+235=z−03=λ{{x - {4 \over 3}} \over { - 1}} = {{y + {2 \over 3}} \over 5} = {{z - 0} \over 3} = \lambda

(say) Let coordinates of foot of perpendicular be

F(−λ+43,5λ−23,3λ)F\left( { - \lambda + {4 \over 3},5\lambda - {2 \over 3},3\lambda } \right)
PF→=(−λ+13)i^+(5λ−83)j^+(3λ)k^\overrightarrow {PF} = \left( { - \lambda + {1 \over 3}} \right)\widehat i + \left( {5\lambda - {8 \over 3}} \right)\widehat j + (3\lambda )\widehat k
PF→.a→=0\overrightarrow {PF} .\overrightarrow a = 0
⇒λ−13+25λ−403+9λ=0\Rightarrow \lambda - {1 \over 3} + 25\lambda {{ - 40} \over 3} + 9\lambda = 0
⇒35λ=413⇒λ=41105\Rightarrow 35\lambda = {{41} \over 3} \Rightarrow \lambda = {{41} \over {105}}

Now,

α=−λ+43,β=5λ−23,γ=3λ\alpha = - \lambda + {4 \over 3},\beta = 5\lambda - {2 \over 3},\gamma = 3\lambda
⇒α+β+γ=7λ+23\Rightarrow \alpha + \beta + \gamma = 7\lambda + {2 \over 3}
=7(41105)+23= 7\left( {{{41} \over {105}}} \right) + {2 \over 3}
=5115= {{51} \over {15}}
⇒35(α+β+γ)=5115×35=119\Rightarrow 35(\alpha + \beta + \gamma ) = {{51} \over {15}} \times 35 = 119
Q88
If the shortest distance between the straight lines 3(x−1)=6(y−2)=2(z−1)3(x - 1) = 6(y - 2) = 2(z - 1) and 4(x−2)=2(y−λ)=(z−3),λ∈R4(x - 2) = 2(y - \lambda ) = (z - 3),\lambda \in R is 138{1 \over {\sqrt {38} }}, then the integral value of λ\lambda is equal to :
A 3
B 2
C 5
D −-1
Correct Answer
Option A
Solution
L1:(x−1)2=(y−2)1=(z−1)3r1→=2i^+j^+3k^{L_1}:{{(x - 1)} \over 2} = {{(y - 2)} \over 1} = {{(z - 1)} \over 3}\overrightarrow {{r_1}} = 2\widehat i + \widehat j + 3\widehat k
L2:(x−2)1=y−λ2=z−34r2→=i^+2j^+4k^{L_2}:{{(x - 2)} \over 1} = {{y - \lambda } \over 2} = {{z - 3} \over 4}\overrightarrow {{r_2}} = \widehat i + 2\widehat j + 4\widehat k

Shortest distance = Projection of

a→{\overrightarrow a }

on

r1→×r2→{\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} }
=∣a→.(r1→×r2→)∣∣r1→×r2→∣= {{\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right|} \over {\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right|}}
∣a→.(r1→×r2→)∣=∣1λ−22213124∣=∣14−5λ∣\left| {\overrightarrow a .\left( {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right)} \right| = \left| \begin{array}{lll}1 & {\lambda - 2} & 2 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array} \right| = \left| {14 - 5\lambda } \right|
∣r1→×r2→∣=38\left| {\overrightarrow {{r_1}} \times \overrightarrow {{r_2}} } \right| = \sqrt {38}

∴\therefore

138=∣14−5λ∣38{1 \over {\sqrt {38} }} = {{\left| {14 - 5\lambda } \right|} \over {\sqrt {38} }}
⇒∣14−5λ∣=1\Rightarrow \left| {14 - 5\lambda } \right| = 1
⇒14−5λ=1\Rightarrow 14 - 5\lambda = 1

or

14−5λ=−114 - 5\lambda = - 1
⇒λ=135\Rightarrow \lambda = {{13} \over 5}

or 3 ∴\therefore Integral value of λ\lambda = 3.

Q89
Let the foot of perpendicular from a point P(1, 2, −-1) to the straight line L:x1=y0=z−1L:{x \over 1} = {y \over 0} = {z \over { - 1}} be N. Let a line be drawn from P parallel to the plane x + y + 2z = 0 which meets L at point Q. If α\alpha is the acute angle between the lines PN and PQ, then cosα\alpha is equal to ________________.
A 15{1 \over {\sqrt 5 }}
B 32{{\sqrt 3 } \over 2}
C 13{1 \over {\sqrt 3 }}
D 123{1 \over {2\sqrt 3 }}
Correct Answer
Option C
Solution
PN→.(i^−k^)=0\overrightarrow {PN} .(\widehat i - \widehat k) = 0

⇒\Rightarrow N(1, 0, −-1) Now,

PQ→.(i^+j^+2k^)=0\overrightarrow {PQ} .(\widehat i + \widehat j + 2\widehat k) = 0

⇒\Rightarrow μ\mu = −- 1 ⇒\Rightarrow Q (−-1, 0, 1)

PN→\overrightarrow {PN}

= 2

j^\widehat j

and

PQ→\overrightarrow {PQ}

=

2i^+2j^−2k^2\widehat i + 2\widehat j - 2\widehat k
⇒cos⁡α=13\Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}
Q90
For real numbers α\alpha and β\beta ≠\ne 0, if the point of intersection of the straight lines x−α1=y−12=z−13{{x - \alpha } \over 1} = {{y - 1} \over 2} = {{z - 1} \over 3} and x−4β=y−63=z−73{{x - 4} \over \beta } = {{y - 6} \over 3} = {{z - 7} \over 3}, lies on the plane x + 2y −- z = 8, then α\alpha −- β\beta is equal to :
A 5
B 9
C 3
D 7
Correct Answer
Option D
Solution

First line is (ϕ\phi + α\alpha, 2ϕ\phi + 1, 3ϕ\phi + 1) and second line is (qβ\beta + 4, 3q + 6, 3q + 7) For intersection ϕ\phi + α\alpha = qβ\beta + 4 ...... (i) 2ϕ\phi + 1 = 3q + 6 .... (ii) 3ϕ\phi + 1 = 3q + 7 ...... (iii) for (ii) & (iii) ϕ\phi = 1, q = −-1 So, from (i) α\alpha + β\beta = 3 Now, point of intersection is (α\alpha + 1, 3, 4) It lies on the plane.

Hence, α\alpha = 5 & β\beta = −-2

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