Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes r.(^i−^j+2^k)=5 and r.(3^i+^j+2^k)=6.
Consider the required line be parallel to vector b given by , b=b1^i+b2^j+b3^k
The position vector of the point (1, 2, 3) is a = ^i+2^j+3^k
The equation of line passing through (1, 2, 3) and parallel to b is given by r = a + λ b
∴ r=(^i+2j+3^k)+λ(b1^i+b2j+b3^k) . . . (i)
The equations of the given planes are
r.(^i−^j+2^k)=5 ⋯(ii)and r.(3^i+^j+^k)=6 ⋯(iii)
The line in Eq. (i) and plane in Eq. (ii) are parallel. Therefore, the normal to the plane of Eq. (ii) and the given line are perpendicular.
∴ (^i−^j+2^k).λ(b1^i+b2^j+b3^k)=0⇒ λ(b1−b2+2b3)=0⇒(b1−b2+2b3)=0 ⋯(iv)
Similarly, (3^i+^j+2^k).λ(b1^i+b2^j+b3^k)=0
⇒ λ(3b1+b2+b3)=0 ⋯(v)
From Eqs. (iv) and (v), we obtain
b1(−1)×1−1×2=b22×3−1×1=b31×1−3(−1)⇒b1−3=b25=b34
Therefore, the direction ratios of b are -3, 5 and 4.
∴ b=b1^i+b2^j+b3^k=−3^i+5^j+4^k
Substituting the value of b in Eq. (i), we obtain
r=(^i+2^j+3^k)+λ(−3^i+5^j+4^k)
This is the equation of the required line .
Note If any line is parallel to the plane, then normal to the plane is perpendicular to the line.