A line 'l' passing through origin is perpendicular to the lines
l_1:\overrightarrow{r} =(3+t)\widehat{i} +(-1+2t)\widehat{j}+ (4+2t)\widehat{k}
l_2:\overrightarrow{r} =(3+2s)\widehat{i} +(3+2s)\widehat{j}+ (2+s)\widehat{k}
If the co-ordinates of the point in the first octant on 'l2' at a distance of \sqrt{17} from the point of intersection of 'l' and 'l1' are (a, b, c), then 18(a + b + c) is equal to ______.
Option: 1 36
Option: 2 54
Option: 3 72
Option: 4 44

Answers (1)

Let equation of line l is

l: \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}

since the line, l pass via origin so,

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

\\\ell_{1}: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}}\\ \ell_{2}: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{\mathrm{i}}+(3+2 \mathrm{~s}) \hat{\mathrm{j}}+(4+\mathrm{s}) \hat{\mathrm{k}}\\

or

\\\ell_{1}: \overrightarrow{\mathrm{r}}=(3\hat i-\hat j+4\hat k)+t(\hat i+2\hat j+2\hat k)\\\ell_{2}: \overrightarrow{\mathrm{r}}=(3\hat i+3\hat j+4\hat k)+s(2\hat i+2\hat j+\hat k)

And also line l is perpendicular to l1 and 12

\\\mathrm{DR} \text { of } \ell_{1} \equiv(1,2,2)\\ \mathrm{DR} \text { of } \ell_{2} \equiv(2,2,1)

\\D R^{\prime} \text { s of } l \text { is }\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{array}\right|=-2 \hat{i}+3 \hat{j}+2 \hat{k}

\\l: \frac{x}{-2}=\frac{y}{3}=\frac{z}{2}=k_1\\\text { Now, } A\left(-2 k_{1}, 3 k_{1},-2 k_{1}\right)\\

Point A lies on lie on the line l1

\\\therefore\left(-2 k_{1}\right) \hat{i}+\left(3 k_{1}\right) \hat{j}-\left(2 k_{1}\right) \hat{k}=(3+t) \hat{i}+(-1+2 t) \hat{j} +(4+2 t) \hat{k} \\ \Rightarrow \quad 3+t=-2 k_{1},-1+2 t=3 k_{1}, 4+2 t=-2 k_{1}\\\Rightarrow k_1=-1\\A(2,-3,2)

\\\text{Let any point on } l_{2}\;\;\text{is }(3+2 s, 3+2 s, 2+s)\\ \sqrt{(2-3-2 s)^{2}(-3-3+2 s)^{2}+(2-2-s)^{2}}=\sqrt{17} \\9 s^{2}+28 s+37=17 \\9 s^{2}+28 s+20=0 \\\Rightarrow \quad 9 s^{2}+18 s+10 s+20=0 \\\Rightarrow (9 s+10)(s+2)=0 \\\therefore s=-2, \frac{-10}{9}

\text{ Hence, }(-1,-1,0) \text{ and } \left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right) \text{are required points.}

Preparation Products

Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
Buy Now
Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
Buy Now
Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
Buy Now
Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-
Buy Now
JEE Main Rank Booster 2021

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, 24x7 Doubt Chat Support,.

₹ 13999/- ₹ 6999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions