# 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

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.

#### 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
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