Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The length of the perpendicular from the point $(2,-1,4)$ on the straight line, $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$ is :

Option 1)
greater than $2$ but less than $3$

Option 2)
greater than $4$

Option 3)
less than $2$

Option 4)
greater than $3$ but less than $4$

Answers (1)

best_answer

$Point=(2,-1,4)=P$

$Line :$ $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$

Length of perpendicular from point to line ?

Any point Q on line $=10\lambda -3,\; -7\lambda+2,\lambda$

DR's of $PQ=\left ( 10\lambda-3-2,-7\lambda+2+1,\lambda-4 \right )$

$=\left ( 10\lambda-5,-7\lambda+3,\lambda-4 \right )$

PQ is perpendicular to given line

$\therefore 10(10\lambda-5)+(-7)(-7\lambda+3)+1(\lambda-4)=0$

$\\=100\lambda-50+49\lambda-21+\lambda-4=0\\\; \; \; \\150\lambda-75=0$

$\lambda=\frac{1}{2}$

$PQ=\sqrt{(10\lambda-5)^{2}+(-7\lambda+3)^{2}+(\lambda-4)^{2}}$

$=\sqrt{(10\times\frac{1}{2}-5)^{2}+(\frac{-7}{2}+3)^{2}+(\frac{1}{2}-4)^{2}}$

$=\sqrt{0+(\frac{-1}{2})^{2}+(\frac{1-8}{2})^{2}}$

$=\sqrt{\frac{1}{4}+\frac{49}{A}}=\sqrt{\frac{50}{4}}=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}}$

$3<\frac{5}{\sqrt{2}}<4$

Option 1)

greater than $2$ but less than $3$

Option 2)

greater than $4$

Option 3)

less than $2$

Option 4)

greater than $3$ but less than $4$

Posted by

solutionqc

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

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs

anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question