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3D Geometry question

2010 · Shift 1 · Q43
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  5. /2010 · Shift 1 · Q43

3D Geometry question

2010 · Shift 1 · Q43

JEE AdvancedMathematics3D GeometryMCQ+4 / −1
Equation of the plane containing the straight line x2=y3=z4{x \over 2} = {y \over 3} = {z \over 4}2x​=3y​=4z​ and perpendicular to the plane containing the straight lines x3=y4=z2{x \over 3} = {y \over 4} = {z \over 2}3x​=4y​=2z​ and x4=y2=z3{x \over 4} = {y \over 2} = {z \over 3}4x​=2y​=3z​ is
  1. A
    x+2y−2z=0x+2y-2z=0x+2y−2z=0
  2. B
    3x+2y−2z=03x+2y-2z=03x+2y−2z=0
  3. C
    x−2y+z=0x-2y+z=0x−2y+z=0
  4. D
    5x+2y−4z=05x+2y-4z=05x+2y−4z=0
View written solutionFree

Correct answer: C

  1. Interpret the given lines as lines through the origin with direction vectors

The line x2=y3=z4\frac{x}{2}=\frac{y}{3}=\frac{z}{4}2x​=3y​=4z​ has direction vector d⃗=(2,3,4).\vec d=(2,3,4).d=(2,3,4).

The two lines x3=y4=z2,x4=y2=z3\frac{x}{3}=\frac{y}{4}=\frac{z}{2}, \qquad \frac{x}{4}=\frac{y}{2}=\frac{z}{3}3x​=4y​=2z​,4x​=2y​=3z​ have direction vectors a⃗=(3,4,2),b⃗=(4,2,3).\vec a=(3,4,2), \qquad \vec b=(4,2,3).a=(3,4,2),b=(4,2,3).

  1. Find the plane containing the two given lines

Since both lines pass through the origin, the plane containing them also passes through the origin. Its normal vector is n⃗1=a⃗×b⃗.\vec n_1=\vec a\times \vec b.n1​=a×b.

Compute:

a⃗×b⃗=∣i^j^k^342423∣\vec a\times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & 4 & 2\\ 4 & 2 & 3 \end{vmatrix}a×b=​i^34​j^​42​k^23​​ =i^(4⋅3−2⋅2)−j^(3⋅3−2⋅4)+k^(3⋅2−4⋅4)=\hat i(4\cdot 3-2\cdot 2)-\hat j(3\cdot 3-2\cdot 4)+\hat k(3\cdot 2-4\cdot 4)=i^(4⋅3−2⋅2)−j^​(3⋅3−2⋅4)+k^(3⋅2−4⋅4) =i^(12−4)−j^(9−8)+k^(6−16)=(8,−1,−10).=\hat i(12-4)-\hat j(9-8)+\hat k(6-16) =(8,-1,-10).=i^(12−4)−j^​(9−8)+k^(6−16)=(8,−1,−10).

So the first plane has normal n⃗1=(8,−1,−10).\vec n_1=(8,-1,-10).n1​=(8,−1,−10).

  1. Let the required plane have normal vector n⃗\vec nn

The required plane:

  • contains the line with direction vector (2,3,4)(2,3,4)(2,3,4), so its normal must be perpendicular to (2,3,4)(2,3,4)(2,3,4),
  • is perpendicular to the plane with normal (8,−1,−10)(8,-1,-10)(8,−1,−10), so its normal must also be perpendicular to (8,−1,−10)(8,-1,-10)(8,−1,−10).

Hence n⃗\vec nn must be perpendicular to both vectors (2,3,4)(2,3,4)(2,3,4) and (8,−1,−10)(8,-1,-10)(8,−1,−10). Therefore, n⃗=(2,3,4)×(8,−1,−10).\vec n=(2,3,4)\times(8,-1,-10).n=(2,3,4)×(8,−1,−10).

Compute:

n⃗=∣i^j^k^2348−1−10∣\vec n= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 3 & 4\\ 8 & -1 & -10 \end{vmatrix}n=​i^28​j^​3−1​k^4−10​​ =i^(3⋅(−10)−4⋅(−1))−j^(2⋅(−10)−4⋅8)+k^(2⋅(−1)−3⋅8)=\hat i(3\cdot(-10)-4\cdot(-1)) -\hat j(2\cdot(-10)-4\cdot 8) +\hat k(2\cdot(-1)-3\cdot 8)=i^(3⋅(−10)−4⋅(−1))−j^​(2⋅(−10)−4⋅8)+k^(2⋅(−1)−3⋅8) =i^(−30+4)−j^(−20−32)+k^(−2−24)=(−26,52,−26).=\hat i(-30+4)-\hat j(-20-32)+\hat k(-2-24) =(-26,52,-26).=i^(−30+4)−j^​(−20−32)+k^(−2−24)=(−26,52,−26).

So, n⃗∝(1,−2,1).\vec n\propto (1,-2,1).n∝(1,−2,1).

  1. Write the equation of the required plane

Since the plane contains the given line, and that line passes through the origin, the plane passes through the origin. Thus its equation is x−2y+z=0.x-2y+z=0.x−2y+z=0.

  1. Check with the options

This matches:

Option C: x−2y+z=0x-2y+z=0x−2y+z=0

  1. Compare with stored correct answer

Stored correct answer is C, which agrees with the derived answer.

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