Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Let the line \frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5} intersect the lines \frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1} and \frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1} at the points A and B respectively. Then the distance of the mid-point of the line segment A B from the plane $2 x-2 y+z=$ 14 is

Option: 1

3


Option: 2

\frac{10}{3}


Option: 3

4


Option: 4

\frac{11}{3}


Answers (1)

best_answer

\begin{aligned} & \frac{x}{1}=\frac{y-6}{-2}=\frac{z+8}{5}=\lambda \, \, \, \, \,\, \, \, \, \, \, \, ...(1)\\ & \frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}=\mu \, \, \, \, ...(2)\\ & \frac{x+3}{6}=\frac{y-3}{-3}=\frac{z-6}{1}=\gamma\, \, \, \, ...(3) \end{aligned}

Intersection of (1) & (2) “A”

\begin{aligned} & (\lambda,-2 \lambda+6,5 \lambda-8) \&(4 \mu+5,3 \mu+7, \mu-2) \\ & \lambda=1, \mu=-1 \\ & A(1,4,-3) \end{aligned}

Intersection (1) & (3) “B”

\begin{aligned} & (\lambda,-2 \lambda+6,5 \lambda-8) \&(6 \gamma-3,-3 \gamma+3, \gamma+6) \\ & \lambda=3 \\ & \gamma=1 \\ & B(3,0,7) \end{aligned}

Mod point of \mathrm{A} \& \mathrm{~B} \Rightarrow(2,2,2)

Perpendicular distance from the plane

\begin{aligned} & 2 x-2 y+z=14 \\ & \left|\frac{2(2)-2(2)+2-14}{\sqrt{4+4+1}}\right|=4 \end{aligned}

Posted by

shivangi.shekhar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question