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Mathematics Question on 3D Geometry

Let L1:r⃗=(i^−j^+2k^)+λ(i^−j^+2k^)L_1 : \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda (\hat{i} - \hat{j} + 2\hat{k}), λ∈R\lambda \in \mathbb{R}
L2:r⃗=(j^−k^)+μ(3i^+j^+pk^)L_2 : \vec{r} = (\hat{j} - \hat{k}) + \mu (3\hat{i} + \hat{j} + p\hat{k}), μ∈R\mu \in \mathbb{R} and
L3:r⃗=δ(ℓi^+mj^+nk^)L_3 : \vec{r} = \delta (\ell \hat{i} + m \hat{j} + n \hat{k}), δ∈R\delta \in \mathbb{R}
Be three lines such that L1L_1 is perpendicular to L2L_2 and L3L_3 is perpendicular to both L1L_1 and L2L_2. Then the point which lies on L3L_3 is

A

(−1,7,4)(-1, 7, 4)

B

(−1,−7,4)(-1, -7, 4)

C

(1,7,−4)(1, 7, -4)

D

(1,−7,4)(1, -7, 4)

Answer

(−1,7,4)(-1, 7, 4)

Explanation

Solution

Identify the direction vectors: L1:d1⃗=i^−j^+2k^L_1: \vec{d_1} = \hat{i} - \hat{j} + 2\hat{k}; L2:d2⃗=3i^+j^+pk^L_2: \vec{d_2} = 3\hat{i} + \hat{j} + p\hat{k}; L3:d3⃗=i^+mj^−k^L_3: \vec{d_3} = \hat{i} + m\hat{j} - \hat{k}

Since L1L_1 is perpendicular to L2L_2, we have:

d1⃗×d2⃗=0\vec{d_1} \times \vec{d_2} = 0.

(1)(3)+(−1)(1)+(2)(p)=0⇒2+2p=0⇒p=−1.(1)(3) + (-1)(1) + (2)(p) = 0 \Rightarrow 2 + 2p = 0 \Rightarrow p = -1.

Since L3L_3 is perpendicular to both L1L_1 and L2L_2: For L3L_3 perpendicular to L1L_1:

d3⃗×d1⃗=0\vec{d_3} \times \vec{d_1} = 0.

(1)(1)+(m)(−1)+(−1)(2)=0⇒−m=1⇒m=−1.(1)(1) + (m)(-1) + (-1)(2) = 0 \Rightarrow -m = 1 \Rightarrow m = -1.

Substitute δ=−1\delta = -1 in r⃗=δ(i^−j^−k^)\vec{r} = \delta(\hat{i} - \hat{j} - \hat{k}) to find the point:

(−1,7,4)(-1, 7, 4).