SATHEE: 3D Geometry 2 Question 9

3D Geometry 2 Question 9

9. A line $l$ passing through the origin is perpendicular to the lines

(2013 Adv.)

$l_{1} : (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} + (4 + 2 t) \hat{k}, - \infty < t < \infty$

$l_{2} : (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}, - \infty < s < \infty$

Then, the coordinate(s) of the point(s) on $l_{2}$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_{1}$ is (are)

(a) $\frac{7}{3}, \frac{7}{3}, \frac{5}{3}$

(b) $(- 1, - 1, 0)$

(d) $\frac{7}{9}, \frac{7}{9}, \frac{8}{9}$

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Answer:

Correct Answer: 9. $(b, d)$

Solution:

Key Idea Equation of straight line is $/ : \frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c}$

Since, $I$ is perpendicular to $I_{1}$ and $I_{2}$ .

So, its DR’s are cross-product of $I_{1}$ and $I_{2}$

Now, to find a point on $I_{2}$ whose distance is given, assume a point and find its distance to obtain point.

Let $l : \frac{x - 0}{a} = \frac{y - 0}{b} = \frac{z - 0}{c}$

which is perpendicular to

$\begin{aligned} l_{1} : (3 \hat{i} - \hat{j} + 4 \hat{k}) + t (\hat{i} + 2 \hat{j} + 2 \hat{k}) \\ l_{2} : (3 \hat{i} + 3 \hat{j} + 2 \hat{k}) + s (2 \hat{i} + 2 \hat{j} + \hat{k}) \\ ∴ & DR’s of l is | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{array} | = - 2 \hat{i} + 3 \hat{j} - 2 \hat{k} \\ l : \frac{x}{- 2} = \frac{y}{3} = \frac{z}{- 2} = k_{1}, k_{2} \end{aligned}$

Now, $A (- 2 k_{1}, 3 k_{1}, - 2 k_{1})$ and $B (- 2 k_{2}, 3 k_{2}, - 2 k_{2})$ .

Since, $A$ lies on $l_{1}$ .

$\begin{aligned} ∴ (- 2 k_{1}) \hat{i} + (3 k_{1}) \hat{j} - (2 k_{1}) \hat{k} = (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} \\ + (4 + 2 t) \hat{k} \\ \Rightarrow 3 + t = - 2 k_{1}, - 1 + 2 t = 3 k_{1}, 4 + 2 t = - 2 k_{1} \\ ∴ k_{1} = - 1 \end{aligned}$

$\Rightarrow$

$A (2, - 3, 2)$

Let any point on $l_{2} (3 + 2 s, 3 + 2 s, 2 + s)$

$\begin{array}{rcrl} \sqrt{(2 - 3 - 2 s)^{2} + (- 3 - 3 - 2 s)^{2} + (2 - 2 - s)^{2}} = \sqrt{17} \\ \Rightarrow & 9 s^{2} + 28 s + 37 = 17 \\ \Rightarrow & 9 s^{2} + 28 s + 20 = 0 \\ \Rightarrow & 9 s^{2} + 18 s + 10 s + 20 = 0 \\ \Rightarrow & (9 s + 10) (s + 2) = 0 \\ ∴ & s & s = - 2, \frac{- 10}{9} . \end{array}$