Question

Find the vector equation of the line passing through(1, 2, 3)and parallel to the planes \(\vec r.=(\hat i-\hat j+2\hat k)=5\) and \(\vec r.(3\hat i+\hat j+\hat k)=6\).

Answer 1

Let the required line be parallel to vector \(\vec b\) given by,
\(\vec b=b_1\hat i+b_2\hat j+b_3\hat k\)

The position vector of the point (1, 2, 3) is
\(\vec a=\hat i+2\hat j+3\hat k\)

The equation of line passing through (1, 2, 3) and parallel to \(\vec b\) is given by,

\(\vec r=\vec a+λ\vec b\)

⇒\(\vec r =\)(\(\hat i+2\hat j+3\hat k\)) + λ(\(b_1\hat i+b_2\hat j+b_3\hat k\))      ...(1)

The equations of the given planes are

\(\vec r.(\hat i-\hat j+2\hat k)=5\)       ...(2)

\(\vec r.(3\hat i+\hat j+\hat k)=6\)       ...(3)

The line in equation (1) and plane in equation (2) are parallel.

Therefore, the normal to the plane of equation (2) and the given line are perpendicular.
\(⇒\)(\(\hat i-\hat j+2\hat k\)) . λ(\(b_1\hat i+b_2\hat j+b_3\hat k\)) = 0
\(⇒λ(b_1-b_2+2b_3)=0\)
\(⇒b_1-b_2+2b_3=0\)     ...(4)

Similarly,
\((3\hat i+\hat j+\hat k).λ(b_1\hat i+b_2\hat j+b_3\hat k^)=0\)
\(⇒λ(3b_1+b_2+b_3)=0\)
\(⇒3b_ 1+b_2+b_3=0\)    ...(5)

From equation (4) and (5), we obtain
\(\frac {b_1}{(-1)×1-1×2}\) = \(\frac {b_2}{2×3-1×1 }\)= \(\frac {b_3}{1×1-3(-1)}\)

\(⇒\frac {b_1}{-3} =\frac {b_2}{5} =\frac {b_3}{4}\)

Therefore, the direction ratios of \(\vec b\) are -3, 5 and 4.

\(∴b=b_1\hat i+b_2\hat j+b_3\hat k\)
\(b =-3\hat i+5\hat j+4\hat k\)

Substituting the values of \(\vec b\) in equation (1), we obtain
\(\vec r=(\hat i+2\hat j+3\hat k)+λ(-3\hat i+5\hat j+4\hat k)\)

This is the equation of the required line.