Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
826 views
in Three Dimensional Geometry by (4.0k points)
closed by

(a) Write the Cartesian equation of the straight line through the points (1, 2, 3) and along the vector 3i + j +2k 

(b) Write a general point on this straight line. 

(c) Find the point of intersection of this straight line with the plane 2x + 3y – z + 2 = 0 

(d) Find the distance from (1, 2, 3) to the plane 2x + 3y – z + 2 = 0

1 Answer

+1 vote
by (3.3k points)
selected by
 
Best answer

(a) Cartesian equation is \(\frac{x-x_{1}} {a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\, \)

\(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\)

(b) Let 

\(\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}\) = λ

x = 3λ + 1, y = λ + 2, z = 2λ + 3 are the general point of the line.

(c) Equation of the plane is 2x + 3y – z + 20….(1)

putting the values of x, y, z in (1) is

2(3λ + 1) – 3(λ + 2) – (2λ + 3) + 2 = 0 

7λ + 7 = 0 ⇒ λ = -1

 ∴ Point of intersection is x = -3 + 1 = -2; y = -1 + 2 = 1, z = -2 + 3 = 1

 ie; (-2, 1, 1)

(d) Distance =

 \(\frac{2(1)+3(2)-3+2}{\sqrt{4+9+1}}\\=\frac{7}{\sqrt{14}}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...