JEE Challenger
More from Three Dimensional Geometry
MathHardAdv 2013 • Paper 1

Coordinates on Line at Fixed Distance from Intersection Point

Let a straight line ll passing through the origin be perpendicular to the two lines l1:(3+t)i^+(−1+2t)j^+(4+2t)k^,−∞<t<∞l_1 : (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}, \quad -\infty < t < \infty l2:(3+2s)i^+(3+2s)j^+(2+s)k^,−∞<s<∞l_2 : (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}, \quad -\infty < s < \infty Then, the coordinate(s) of the point(s) lying on l2l_2 at a distance of 17\sqrt{17} from the point of intersection of ll and l1l_1 is (are)

Options

A

(73,73,53)\left(\frac{7}{3}, \frac{7}{3}, \frac{5}{3}\right)

B

(−1,−1,0)(-1, -1, 0)

Correct
C

(1,1,1)(1, 1, 1)

D

(79,79,89)\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)

Correct

Topics & Concepts

Three Dimensional GeometryLine in 3D

Step-by-Step Solution

To find the coordinates of the required point(s) on line l2l_2, we follow a step-by-step approach.

Step 1: Determine the direction vector of line ll

The given lines l1l_1 and l2l_2 in vector form are: l1:r⃗1(t)=(3i^−j^+4k^)+t(i^+2j^+2k^)l_1 : \vec{r}_1(t) = (3\hat{i} - \hat{j} + 4\hat{k}) + t(\hat{i} + 2\hat{j} + 2\hat{k}) l2:r⃗2(s)=(3i^+3j^+2k^)+s(2i^+2j^+k^)l_2 : \vec{r}_2(s) = (3\hat{i} + 3\hat{j} + 2\hat{k}) + s(2\hat{i} + 2\hat{j} + \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^+2j^+k^\vec{d}_2 = 2\hat{i} + 2\hat{j} + \hat{k}

Since line ll is perpendicular to both l1l_1 and l2l_2, its direction vector d⃗\vec{d} is parallel to d⃗1×d⃗2\vec{d}_1 \times \vec{d}_2:

d⃗=d⃗1×d⃗2=∣i^j^k^122221∣=(2−4)i^−(1−4)j^+(2−4)k^=−2i^+3j^−2k^\vec{d} = \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = (2 - 4)\hat{i} - (1 - 4)\hat{j} + (2 - 4)\hat{k} = -2\hat{i} + 3\hat{j} - 2\hat{k}

Since ll passes through the origin (0,0,0)(0,0,0), the parametric equation of line ll is given by: l:r⃗l(u)=−2ui^+3uj^−2uk^,−∞<u<∞l : \vec{r}_l(u) = -2u\hat{i} + 3u\hat{j} - 2u\hat{k}, \quad -\infty < u < \infty

Step 2: Find the point of intersection of ll and l1l_1

Let PP be the point of intersection of line ll and line l1l_1. Equating the position vectors for ll and l1l_1: 3+t=−2u— (1)3 + t = -2u \quad \text{--- (1)} −1+2t=3u— (2)-1 + 2t = 3u \quad \text{--- (2)} 4+2t=−2u— (3)4 + 2t = -2u \quad \text{--- (3)}

From equations (1) and (3): 3+t=4+2t  ⟹  t=−13 + t = 4 + 2t \implies t = -1

Substituting t=−1t = -1 into equation (1): 3+(−1)=−2u  ⟹  u=−13 + (-1) = -2u \implies u = -1

Checking in equation (2): −1+2(−1)=−3=3(−1)-1 + 2(-1) = -3 = 3(-1) This satisfies all equations. Thus, substituting u=−1u = -1 into the parametric equation of ll, we get the coordinates of point PP: P=(2,−3,2)P = (2, -3, 2)

Step 3: Find the point(s) on l2l_2 at distance 17\sqrt{17} from PP

Any general point QQ on line l2l_2 has coordinates: Q(s)=(3+2s,3+2s,2+s)Q(s) = (3 + 2s, 3 + 2s, 2 + s)

The distance between P(2,−3,2)P(2, -3, 2) and Q(s)Q(s) is given to be 17\sqrt{17}, so PQ2=17PQ^2 = 17: [(3+2s)−2]2+[(3+2s)−(−3)]2+[(2+s)−2]2=17[(3 + 2s) - 2]^2 + [(3 + 2s) - (-3)]^2 + [(2 + s) - 2]^2 = 17 (1+2s)2+(6+2s)2+s2=17(1 + 2s)^2 + (6 + 2s)^2 + s^2 = 17

Expanding the terms: (1+4s+4s2)+(36+24s+4s2)+s2=17(1 + 4s + 4s^2) + (36 + 24s + 4s^2) + s^2 = 17 9s2+28s+37=179s^2 + 28s + 37 = 17 9s2+28s+20=09s^2 + 28s + 20 = 0

Factorizing the quadratic equation: 9s2+18s+10s+20=09s^2 + 18s + 10s + 20 = 0 9s(s+2)+10(s+2)=09s(s + 2) + 10(s + 2) = 0 (9s+10)(s+2)=0(9s + 10)(s + 2) = 0

So, the values of ss are: s=−2ors=−109s = -2 \quad \text{or} \quad s = -\frac{10}{9}

Step 4: Calculate coordinates corresponding to each value of ss

  1. For s=−2s = -2: Q=(3+2(−2),3+2(−2),2+(−2))=(−1,−1,0)Q = (3 + 2(-2), 3 + 2(-2), 2 + (-2)) = (-1, -1, 0)

  2. For s=−109s = -\frac{10}{9}: Q=(3+2(−109),3+2(−109),2+(−109))=(79,79,89)Q = \left(3 + 2\left(-\frac{10}{9}\right), 3 + 2\left(-\frac{10}{9}\right), 2 + \left(-\frac{10}{9}\right)\right) = \left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)

Conclusion

The possible coordinates lying on l2l_2 are (−1,−1,0)(-1, -1, 0) and (79,79,89)\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right).

Correct Options: B and D

PreviousNext