A line passes through (2, -1, 3) and is perpendicular to the lines →r=(^i+^j+^k)+λ(2^i+−2^j+^k) and →r=(2^i−^j−3^k)+μ(^i+2^j+2^k). Obtain its equation in vector and Certesian form.
Given line are →r=(^i+^j+^k)+λ(2^i+−2^j+^k) and →r=(2^i−^j−3^k)+μ(^i+2^j+2^k)
A line perpendicular to the given lines will be in the direction of
(2^i−2^j+^k)×(^i+2^j+2^k)=∣∣ ∣ ∣∣^i^j^k2−21122∣∣ ∣ ∣∣=−6^i−3^j+6^k or,→b=2^i+^j−2^k
Position vector of given point (2, -1, 3) is. →a=2^i−^j+3^k
Using →r=→a+λ¯b,the required vector equaton of line is:→r=2hati−^j+3^k+λ(2^i+2^j−2^k)
And, Cartesian form of the line is : x−22=y+11=z−3−2