A line \mathcal{l} passing through the origin is perpendicular to the lines ℓ_{1} : \left(\right. 3 + t \left.\right) \textrm{ } \hat{i} \textrm{ } + \textrm{ } \left(\right. - 1 + 2 t \left.\right) \textrm{ } \hat{j} \textrm{ } + \textrm{ } \left(\right. 4 + 2 t \left.\right) \textrm{ } \hat{k} , - \infty < t < \infty ℓ_{2} : \left(\right. 3 + 2 s \left.\right) \textrm{ } \hat{i} \textrm{ } + \textrm{ } \left(\right. 3 + 2 s \left.\right) \textrm{ } \hat{j} \textrm{ } + \textrm{ } \left(\right. 2 + s \left.\right) \textrm{ } \hat{k} , - \infty < s < \infty Then, the coordinate(s) of the point(s) on \left(\mathcal{l}\right)_{2} at a distance of \sqrt{17} from the point of intersection of l and \left(\mathcal{l}\right)

Engineering

Mathematics

Equation of Straight Line in 3D

Question

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

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

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

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)

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

(1, 1, 1)

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

(– 1, – 1, 0)

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$ℓ_{1} : \frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 4}{2} = λ$

$ℓ_{2} : \frac{x - 3}{2} = \frac{y - 3}{2} = \frac{z - 2}{1} = s$

$∴$ line $l$ is parallel to vector

$= | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{matrix} |$

$\Rightarrow ℓ : \frac{x - 0}{2} = \frac{y - 0}{3} = \frac{z - 0}{- 2} = μ$

Any point on $l$ 1 is ( $λ$ + 3, 2 $λ$ – 1, 2 $λ$ + 4)

and Any point on $l$ is (–2µ, 3µ, – 2µ)

$∴$ on solving $l_{1}$ and $l$ ,

we get (2, – 3, 2)

Now, any point on $l_{2}$ is

(2s + 3, 2s + 3, s + 2)

No, (2s + 1)² + (2s + 6)² + s² = 17

$∴$ for s = – 2, we get (–1, – 1, 0) and for $s = \frac{- 10}{9}, we get (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$

Please subscribe our Youtube channel to unlock this solution.