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3D Geometry question

2025 · 7 Apr · Shift 2 · Q35
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3D Geometry question

2025 · 7 Apr · Shift 2 · Q35

JEE MainMathematics3D GeometryMCQ+4 / −1
If the equation of the line passing through the point (0,−12,0)\left( 0, -\frac{1}{2}, 0 \right)(0,−21​,0) and perpendicular to the lines r⃗=λ(i^+aj^+bk^)\vec{r} = \lambda \left( \hat{i} + a\hat{j} + b\hat{k} \right)r=λ(i^+aj^​+bk^) and r⃗=(i^−j^−6k^)+μ(−bi^+aj^+5k^)\vec{r} = \left( \hat{i} - \hat{j} - 6\hat{k} \right) + \mu \left( -b \hat{i} + a\hat{j} + 5\hat{k} \right)r=(i^−j^​−6k^)+μ(−bi^+aj^​+5k^) is x−1−2=y+4d=z−c−4\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}−2x−1​=dy+4​=−4z−c​, then a+b+c+da+b+c+da+b+c+d is equal to :
  1. A
    13
  2. B
    14
  3. C
    12
  4. D
    10
View written solutionFree

Correct answer: B

  1. Identify the direction vectors of the given lines

The first line is

r⃗=λ(i^+aj^+bk^)\vec r = \lambda(\hat i + a\hat j + b\hat k)r=λ(i^+aj^​+bk^)

so its direction vector is

v⃗1=(1,a,b).\vec v_1 = (1,a,b).v1​=(1,a,b).

The second line is

r⃗=(1,−1,−6)+μ(−b,a,5)\vec r = (1,-1,-6) + \mu(-b,a,5)r=(1,−1,−6)+μ(−b,a,5)

so its direction vector is

v⃗2=(−b,a,5).\vec v_2 = (-b,a,5).v2​=(−b,a,5).

The required line is perpendicular to both these lines, so its direction vector must be parallel to

v⃗1×v⃗2.\vec v_1 \times \vec v_2.v1​×v2​.

Also, from

x−1−2=y+4d=z−c−4,\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4},−2x−1​=dy+4​=−4z−c​,

the direction vector of the required line is

(−2,d,−4).(-2,d,-4).(−2,d,−4).
  1. Use the fact that the required line passes through (0,−12,0)\left(0,-\frac12,0\right)(0,−21​,0)

A symmetric line

x−1−2=y+4d=z−c−4=t\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}=t−2x−1​=dy+4​=−4z−c​=t

can be written as

x=1−2t,y=−4+dt,z=c−4t.x=1-2t,\qquad y=-4+dt,\qquad z=c-4t.x=1−2t,y=−4+dt,z=c−4t.

Since it passes through (0,−12,0)\left(0,-\frac12,0\right)(0,−21​,0), substitute:

  • From x=0x=0x=0: 0=1−2t  ⟹  t=12.0=1-2t \implies t=\frac12.0=1−2t⟹t=21​.

  • From z=0z=0z=0: 0=c−4(12)=c−2  ⟹  c=2.0=c-4\left(\frac12\right)=c-2 \implies c=2.0=c−4(21​)=c−2⟹c=2.

  • From y=−12y=-\frac12y=−21​: −12=−4+d(12).-\frac12=-4+d\left(\frac12\right).−21​=−4+d(21​). So, −12+4=d2  ⟹  72=d2  ⟹  d=7.-\frac12+4=\frac d2 \implies \frac72=\frac d2 \implies d=7.−21​+4=2d​⟹27​=2d​⟹d=7.

Thus,

c=2,d=7.c=2,\qquad d=7.c=2,d=7.
  1. Find aaa and bbb using perpendicularity

Since the required direction vector (−2,7,−4)(-2,7,-4)(−2,7,−4) is perpendicular to both given lines,

With v⃗1=(1,a,b)\vec v_1=(1,a,b)v1​=(1,a,b):

(−2,7,−4)⋅(1,a,b)=0(-2,7,-4)\cdot(1,a,b)=0(−2,7,−4)⋅(1,a,b)=0 −2+7a−4b=0-2+7a-4b=0−2+7a−4b=0 7a−4b=2.(1)7a-4b=2. \qquad (1)7a−4b=2.(1)

With v⃗2=(−b,a,5)\vec v_2=(-b,a,5)v2​=(−b,a,5):

(−2,7,−4)⋅(−b,a,5)=0(-2,7,-4)\cdot(-b,a,5)=0(−2,7,−4)⋅(−b,a,5)=0 2b+7a−20=02b+7a-20=02b+7a−20=0 7a+2b=20.(2)7a+2b=20. \qquad (2)7a+2b=20.(2)

Now solve (1) and (2):

From (2) −-− (1):

(7a+2b)−(7a−4b)=20−2(7a+2b)-(7a-4b)=20-2(7a+2b)−(7a−4b)=20−2 6b=18  ⟹  b=3.6b=18 \implies b=3.6b=18⟹b=3.

Substitute into (2):

7a+2(3)=207a+2(3)=207a+2(3)=20 7a+6=207a+6=207a+6=20 7a=14  ⟹  a=2.7a=14 \implies a=2.7a=14⟹a=2.

So,

a=2,b=3.a=2,\qquad b=3.a=2,b=3.
  1. Compute a+b+c+da+b+c+da+b+c+d
a+b+c+d=2+3+2+7=14.a+b+c+d=2+3+2+7=14.a+b+c+d=2+3+2+7=14.
  1. Match with the options

The correct value is

14.14.14.

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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