If the distance between the plane Ax−2y+z=d and the plane containing the lines x−12=y−23=z−34 and x−23=y−34=z−45 is 6, then the value of |d| is

2l+3m+4n=0l−1=m2=n−1Equation of plane will bea(x−1)+b(y−2)+c(z−3)=0⇒−l(x−1)+2(y−2)−1(z−3)=0⇒−x+1+2y−4−z+3=0⇒−x+2y−z=0⇒x−2y+z=0∴|d|6=6

If the distance between the plane Ax−2y+z=d and the plane containing the lines x−12=y−23=z−34 and x−23=y−34=z−45 is 6, then the value of |d| is

2l+3m+4n=0l−1=m2=n−1Equation of plane will bea(x−1)+b(y−2)+c(z−3)=0⇒−l(x−1)+2(y−2)−1(z−3)=0⇒−x+1+2y−4−z+3=0⇒−x+2y−z=0⇒x−2y+z=0∴|d|6=6

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$If the distance between the plane A x - 2 y + z = d and the plane containing the lines \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and$
$\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5} is \sqrt{6}, then the value of | d | is$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{matrix} 2 l + 3 m + 4 n = 0 \\ \frac{l}{- 1} = \frac{m}{2} = \frac{n}{- 1} \end{matrix}$

Equation of plane will be

$\begin{array}{l} a (x - 1) + b (y - 2) + c (z - 3) = 0 \\ \Rightarrow - l (x - 1) + 2 (y - 2) - 1 (z - 3) = 0 \\ \Rightarrow - x + 1 + 2 y - 4 - z + 3 = 0 \\ \Rightarrow - x + 2 y - z = 0 \\ \Rightarrow x - 2 y + z = 0 \\ ∴ \frac{| d |}{\sqrt{6}} = \sqrt{6} \end{array}$