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3D Geometry question

2007 · Shift 0 · Q62
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3D Geometry question

2007 · Shift 0 · Q62

JEE MainMathematics3D GeometryMCQ+4 / −1
Let LLL be the line of intersection of the planes 2x+3y+z=12x+3y+z=12x+3y+z=1 and x+3y+2z=2.x+3y+2z=2.x+3y+2z=2. If LLL makes an angle α\alphaα with the positive xxx-axis, then cos α\alphaα equals
  1. A
    111
  2. B
    12{1 \over {\sqrt 2 }}2​1​
  3. C
    13{1 \over {\sqrt 3 }}3​1​
  4. D
    12{1 \over 2}21​
View written solutionFree

Correct answer: C

  1. The line LLL is the intersection of the planes 2x+3y+z=12x+3y+z=12x+3y+z=1 and x+3y+2z=2.x+3y+2z=2.x+3y+2z=2.

  2. A direction vector of the line of intersection is perpendicular to the normals of both planes.

    • Normal to the first plane: n⃗1=(2,3,1)\vec n_1=(2,3,1)n1​=(2,3,1)
    • Normal to the second plane: n⃗2=(1,3,2)\vec n_2=(1,3,2)n2​=(1,3,2)

    So, a direction vector of LLL is d⃗=n⃗1×n⃗2.\vec d=\vec n_1\times \vec n_2.d=n1​×n2​.

  3. Compute the cross product:

    \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix}$$ $$\vec d=\hat i(3\cdot 2-1\cdot 3)-\hat j(2\cdot 2-1\cdot 1)+\hat k(2\cdot 3-3\cdot 1)$$ $$\vec d=\hat i(6-3)-\hat j(4-1)+\hat k(6-3)$$ $$\vec d=(3,-3,3)=3(1,-1,1).$$ Thus, we can take the direction vector as $$\vec d=(1,-1,1).$$

  4. If α\alphaα is the angle made by the line with the positive xxx-axis, then cos⁡α=x-component of direction vectormagnitude of direction vector.\cos\alpha=\frac{\text{$x$-component of direction vector}}{\text{magnitude of direction vector}}.cosα=magnitude of direction vectorx-component of direction vector​.

    Using d⃗=(1,−1,1)\vec d=(1,-1,1)d=(1,−1,1), ∣d⃗∣=12+(−1)2+12=3.|\vec d|=\sqrt{1^2+(-1)^2+1^2}=\sqrt{3}.∣d∣=12+(−1)2+12​=3​.

    Therefore, cos⁡α=13.\cos\alpha=\frac{1}{\sqrt{3}}.cosα=3​1​.

  5. Hence the correct option is C 13.\boxed{\text{C } \frac{1}{\sqrt{3}}}.C 3​1​​.

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