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.