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Question

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.

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Solution

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(b1b2+2b3)=0(b1b2+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)×11×2=b22×31×1=b31×13(1)b13=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.


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