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Question
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(x−2)+B(y+1)+C(z−3)=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.
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