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3D Geometry question

2024 · 30 Jan · Shift 2 · Q41
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  5. /2024 · 30 Jan · Shift 2 · Q41

3D Geometry question

2024 · 30 Jan · Shift 2 · Q41

JEE MainMathematics3D GeometryMCQ+4 / −1
Let L1:r⃗=(i^−j^+2k^)+λ(i^−j^+2k^),λ∈RL_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}L1​:r=(i^−j^​+2k^)+λ(i^−j^​+2k^),λ∈R, L2:r⃗=(j^−k^)+μ(3i^+j^+pk^),μ∈R, and L3:r⃗=δ(ℓi^+mj^+nk^),δ∈RL_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text {, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}L2​:r=(j^​−k^)+μ(3i^+j^​+pk^),μ∈R, and L3​:r=δ(ℓi^+mj^​+nk^),δ∈R be three lines such that L1L_1L1​ is perpendicular to L2L_2L2​ and L3L_3L3​ is perpendicular to both L1L_1L1​ and L2L_2L2​. Then, the point which lies on L3L_3L3​ is
  1. A
    (1,7,−4)(1,7,-4)(1,7,−4)
  2. B
    (1,−7,4)(1,-7,4)(1,−7,4)
  3. C
    (−1,7,4)(-1,7,4)(−1,7,4)
  4. D
    (−,1−7,4)(-, 1-7,4)(−,1−7,4)
View written solutionFree

Correct answer: C

  1. Write the given lines in coordinate form

    \vec r=(1,-1,2)+\lambda(1,-1,2)$$ So the direction vector of $L_1$ is $$\vec d_1=(1,-1,2).$$ $$L_2:\vec r=(0,1,-1)+\mu(3,1,p)$$ So the direction vector of $L_2$ is $$\vec d_2=(3,1,p).$$ $$L_3:\vec r=\delta(\ell,m,n)$$ Hence $L_3$ passes through the origin and has direction vector $$\vec d_3=(\ell,m,n).$$

  2. Use the condition L1⊥L2L_1 \perp L_2L1​⊥L2​

    Since perpendicular lines have perpendicular direction vectors, d⃗1⋅d⃗2=0.\vec d_1\cdot \vec d_2=0.d1​⋅d2​=0.

    Therefore, (1,−1,2)⋅(3,1,p)=0(1,-1,2)\cdot(3,1,p)=0(1,−1,2)⋅(3,1,p)=0 1⋅3+(−1)⋅1+2p=01\cdot 3+(-1)\cdot 1+2p=01⋅3+(−1)⋅1+2p=0 3−1+2p=03-1+2p=03−1+2p=0 2+2p=02+2p=02+2p=0 p=−1.p=-1.p=−1.

    So, d⃗2=(3,1,−1).\vec d_2=(3,1,-1).d2​=(3,1,−1).

  3. Use the condition that L3L_3L3​ is perpendicular to both L1L_1L1​ and L2L_2L2​

    Therefore, the direction vector of L3L_3L3​ must be perpendicular to both d⃗1\vec d_1d1​ and d⃗2\vec d_2d2​.

    Hence, d⃗3∥d⃗1×d⃗2.\vec d_3 \parallel \vec d_1\times \vec d_2.d3​∥d1​×d2​.

    Compute the cross product:

    \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & -1 & 2\\ 3 & 1 & -1 \end{vmatrix}$$ $$=\hat i\big((-1)(-1)-2\cdot 1\big)-\hat j\big(1\cdot(-1)-2\cdot 3\big)+\hat k\big(1\cdot 1-(-1)\cdot 3\big)$$ $$=\hat i(1-2)-\hat j(-1-6)+\hat k(1+3)$$ $$=-\hat i+7\hat j+4\hat k.$$ So a direction vector of $L_3$ is $$(-1,7,4).$$

  4. Find a point on L3L_3L3​

    Since L3:r⃗=δ(−1,7,4),L_3:\vec r=\delta(-1,7,4),L3​:r=δ(−1,7,4), any point on L3L_3L3​ is of the form (−δ,7δ,4δ).(-\delta,7\delta,4\delta).(−δ,7δ,4δ).

    Taking δ=1\delta=1δ=1, we get the point (−1,7,4).(-1,7,4).(−1,7,4).

  5. Match with the options

    The point (−1,7,4)(-1,7,4)(−1,7,4) corresponds to Option C.

  6. Comparison with stored answer

    Stored correct answer: C

    Our derived answer: C

    Hence, they agree.

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