JEE Challenger
More from Three Dimensional Geometry
MathHardAdv 2019 • Paper 1

Find Line Perpendicular to Two Given Lines

Suppose L1L_1 and L2L_2 represent the lines r⃗=i^+λ(−i^+2j^+2k^),λ∈Rand\vec{r} = \hat{i} + \lambda(-\hat{i} + 2\hat{j} + 2\hat{k}), \quad \lambda \in \mathbb{R} \quad \text{and} r⃗=μ(2i^−j^+2k^),μ∈R\vec{r} = \mu(2\hat{i} - \hat{j} + 2\hat{k}), \quad \mu \in \mathbb{R} respectively. If L3L_3 is a line that is perpendicular to both L1L_1 and L2L_2 and intersects both of them, then which of the following options represent(s) L3L_3?

Options

A

r⃗=29(4i^+j^+k^)+t(2i^+2j^−k^),t∈R\vec{r} = \frac{2}{9}(4\hat{i} + \hat{j} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), \quad t \in \mathbb{R}

Correct
B

r⃗=29(2i^−j^+2k^)+t(2i^+2j^−k^),t∈R\vec{r} = \frac{2}{9}(2\hat{i} - \hat{j} + 2\hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), \quad t \in \mathbb{R}

Correct
C

r⃗=13(2i^+k^)+t(2i^+2j^−k^),t∈R\vec{r} = \frac{1}{3}(2\hat{i} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}), \quad t \in \mathbb{R}

Correct
D

r⃗=t(2i^+2j^−k^),t∈R\vec{r} = t(2\hat{i} + 2\hat{j} - \hat{k}), \quad t \in \mathbb{R}

Topics & Concepts

Three Dimensional GeometryLine in 3D

Step-by-Step Solution

To find the equation of the line L3L_3 that is perpendicular to both L1L_1 and L2L_2 and intersects both of them, we follow a step-by-step approach.

Step 1: Find the direction vector of L3L_3

The given lines are: L1:r⃗=i^+λ(−i^+2j^+2k^)L_1: \vec{r} = \hat{i} + \lambda(-\hat{i} + 2\hat{j} + 2\hat{k}) L2:r⃗=μ(2i^−j^+2k^)L_2: \vec{r} = \mu(2\hat{i} - \hat{j} + 2\hat{k})

The direction vectors of L1L_1 and L2L_2 are: d⃗1=−i^+2j^+2k^\vec{d}_1 = -\hat{i} + 2\hat{j} + 2\hat{k} d⃗2=2i^−j^+2k^\vec{d}_2 = 2\hat{i} - \hat{j} + 2\hat{k}

Since L3L_3 is perpendicular to both L1L_1 and L2L_2, its direction vector d⃗3\vec{d}_3 is parallel to d⃗1×d⃗2\vec{d}_1 \times \vec{d}_2:

d⃗3=d⃗1×d⃗2=∣i^j^k^−1222−12∣\vec{d}_3 = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 2 & -1 & 2 \end{vmatrix}

d⃗3=i^(4−(−2))−j^(−2−4)+k^(1−4)=6i^+6j^−3k^=3(2i^+2j^−k^)\vec{d}_3 = \hat{i}(4 - (-2)) - \hat{j}(-2 - 4) + \hat{k}(1 - 4) = 6\hat{i} + 6\hat{j} - 3\hat{k} = 3(2\hat{i} + 2\hat{j} - \hat{k})

Thus, the direction of L3L_3 is given by the vector d⃗=2i^+2j^−k^\vec{d} = 2\hat{i} + 2\hat{j} - \hat{k}.

Step 2: Find the points of intersection on L1L_1 and L2L_2

Let PP be a point on L1L_1 and QQ be a point on L2L_2 such that the line segment PQPQ lies along L3L_3: P=(1−λ,2λ,2λ)P = (1-\lambda, 2\lambda, 2\lambda) Q=(2μ,−μ,2μ)Q = (2\mu, -\mu, 2\mu)

The vector PQ⃗\vec{PQ} is: PQ⃗=(2μ+λ−1)i^+(−μ−2λ)j^+(2μ−2λ)k^\vec{PQ} = (2\mu + \lambda - 1)\hat{i} + (-\mu - 2\lambda)\hat{j} + (2\mu - 2\lambda)\hat{k}

Since PQ⊥L1PQ \perp L_1, we have PQ⃗⋅d⃗1=0\vec{PQ} \cdot \vec{d}_1 = 0: −(2μ+λ−1)+2(−μ−2λ)+2(2μ−2λ)=0-(2\mu + \lambda - 1) + 2(-\mu - 2\lambda) + 2(2\mu - 2\lambda) = 0 −2μ−λ+1−2μ−4λ+4μ−4λ=0  ⟹  −9λ+1=0  ⟹  λ=19-2\mu - \lambda + 1 - 2\mu - 4\lambda + 4\mu - 4\lambda = 0 \implies -9\lambda + 1 = 0 \implies \lambda = \frac{1}{9}

Since PQ⊥L2PQ \perp L_2, we have PQ⃗⋅d⃗2=0\vec{PQ} \cdot \vec{d}_2 = 0: 2(2μ+λ−1)−(−μ−2λ)+2(2μ−2λ)=02(2\mu + \lambda - 1) - (-\mu - 2\lambda) + 2(2\mu - 2\lambda) = 0 4μ+2λ−2+μ+2λ+4μ−4λ=0  ⟹  9μ−2=0  ⟹  μ=294\mu + 2\lambda - 2 + \mu + 2\lambda + 4\mu - 4\lambda = 0 \implies 9\mu - 2 = 0 \implies \mu = \frac{2}{9}

Substituting λ=19\lambda = \frac{1}{9} and μ=29\mu = \frac{2}{9} into the coordinates of PP and QQ: r⃗P=(1−19)i^+29j^+29k^=29(4i^+j^+k^)\vec{r}_P = \left(1 - \frac{1}{9}\right)\hat{i} + \frac{2}{9}\hat{j} + \frac{2}{9}\hat{k} = \frac{2}{9}(4\hat{i} + \hat{j} + \hat{k}) r⃗Q=2(29)i^−29j^+2(29)k^=29(2i^−j^+2k^)\vec{r}_Q = 2\left(\frac{2}{9}\right)\hat{i} - \frac{2}{9}\hat{j} + 2\left(\frac{2}{9}\right)\hat{k} = \frac{2}{9}(2\hat{i} - \hat{j} + 2\hat{k})

Step 3: Check the Given Options

Any line equation for L3L_3 can be written as: r⃗=r⃗0+t(2i^+2j^−k^),t∈R\vec{r} = \vec{r}_0 + t(2\hat{i} + 2\hat{j} - \hat{k}), \quad t \in \mathbb{R} where r⃗0\vec{r}_0 is any point lying on L3L_3.

  1. Option (A): r⃗=29(4i^+j^+k^)+t(2i^+2j^−k^)\vec{r} = \frac{2}{9}(4\hat{i} + \hat{j} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}) Here, r⃗0=r⃗P\vec{r}_0 = \vec{r}_P, which is the point of intersection on L1L_1. Thus, Option (A) is correct.

  2. Option (B): r⃗=29(2i^−j^+2k^)+t(2i^+2j^−k^)\vec{r} = \frac{2}{9}(2\hat{i} - \hat{j} + 2\hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}) Here, r⃗0=r⃗Q\vec{r}_0 = \vec{r}_Q, which is the point of intersection on L2L_2. Thus, Option (B) is correct.

  3. Option (C): r⃗=13(2i^+k^)+t(2i^+2j^−k^)\vec{r} = \frac{1}{3}(2\hat{i} + \hat{k}) + t(2\hat{i} + 2\hat{j} - \hat{k}) Let us check if the point r⃗0=13(2i^+k^)=(23,0,13)\vec{r}_0 = \frac{1}{3}(2\hat{i} + \hat{k}) = \left(\frac{2}{3}, 0, \frac{1}{3}\right) lies on L3L_3: (69,0,39)=r⃗P+t(2i^+2j^−k^)=(89+2t,29+2t,29−t)\left(\frac{6}{9}, 0, \frac{3}{9}\right) = \vec{r}_P + t(2\hat{i} + 2\hat{j} - \hat{k}) = \left(\frac{8}{9} + 2t, \frac{2}{9} + 2t, \frac{2}{9} - t\right)

    Equating components:

    • xx-component: 89+2t=69  ⟹  t=−19\frac{8}{9} + 2t = \frac{6}{9} \implies t = -\frac{1}{9}
    • yy-component: 29+2(−19)=0\frac{2}{9} + 2\left(-\frac{1}{9}\right) = 0
    • zz-component: 29−(−19)=39=13\frac{2}{9} - \left(-\frac{1}{9}\right) = \frac{3}{9} = \frac{1}{3}

    Since t=−19t = -\frac{1}{9} satisfies all three coordinates, the point 13(2i^+k^)\frac{1}{3}(2\hat{i} + \hat{k}) lies on L3L_3. Thus, Option (C) is correct.

  4. Option (D): r⃗=t(2i^+2j^−k^)\vec{r} = t(2\hat{i} + 2\hat{j} - \hat{k}) Checking if the origin (0,0,0)(0,0,0) lies on L3L_3: 89+2t=0  ⟹  t=−49\frac{8}{9} + 2t = 0 \implies t = -\frac{4}{9} Substituting t=−49t = -\frac{4}{9} into the yy-component gives 29+2(−49)=−69≠0\frac{2}{9} + 2\left(-\frac{4}{9}\right) = -\frac{6}{9} \neq 0. Hence, the origin does not lie on L3L_3. Thus, Option (D) is incorrect.

Conclusion

The options that correctly represent the line L3L_3 are (A), (B), and (C).

PreviousNext