Advertisement Remove all ads

A line passes through (2, −1, 3) and is perpendicular to the lines r=(i+j-k)+lambda(2i-2j+k) and vecr=(2i-j-3k)+mu(i+2j+2k) . Obtain its equation in vector and Cartesian from. - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
 

A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from. 

 
Advertisement Remove all ads

Solution

Since the line passes through a point (2,-1,3), its equation will be

A(x2)+B(y+1)+C(z3)=0

The normal vector of this line is `Ahati+Bhatj+Chatk.`

Since this line is perpendicular to the two given lines, the dot product of their respective normals will be 0.

`(Ahati+Bhatj+Chatk).(2hati−2hatj+hatk)=0`

`⇒2A−2B+C=0                   .........(1)`

`(Ahati+Bhatj+Chatk).(hati+2hatj+2hatk)=0`

`⇒A+2B+2C=0                   .........(2)`

Adding (1) and (2), we get:

C=A;

Substituing in (1), we get `B=A/2.`
Substituting the values of B and C in the equation of the line, we get:

`A(x−2)+A/2(y+1)−A(z−3)=0`

`⇒2(x−2)+(y+1)−2(z−3)=0`

`⇒2x+y−2z+3=0`

is the required equation of the line.

Concept: Equation of a Line in Space
  Is there an error in this question or solution?
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×