JEE Mathematics 2023 Question with Solution

Given: Plane $P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ and line $L: \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}$. The plane is parallel to the line, and the $y$-intercept of $P$ is $1$.

Find: The distance between plane $P$ and line $L$.

Since $P$ is parallel to $L$, the direction ratios of the line are perpendicular to the normal vector of the plane. Hence,

so,

Also, the $y$-intercept of the plane is $1$. Therefore, putting $x=0, z=0, y=1$ in the plane equation gives

Using this in

we get

Hence the plane becomes

A point on line $L$ is obtained by taking the common parameter $=0$, giving

Since the line is parallel to the plane, the distance between the line and the plane equals the perpendicular distance of any point on the line from the plane.

Therefore,

Therefore, the correct option is D. The solution concludes option D, but the listed options place $\sqrt{14}$ as A. The value obtained from the working is $\sqrt{14}$.

Given: $P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ and $L: \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}$.

Find: Distance between the plane and the line.

For a plane $Ax+By+Cz+D=0$ to be parallel to a line with direction ratios $(l,m,n)$, we must have

Here, normal vector of the plane is $(8, \alpha_1, \alpha_2)$ and direction ratios of the line are $(2,3,5)$. So,

Now use the $y$-intercept condition. On the $y$-axis, $x=0$ and $z=0$. If the intercept is $1$, the point $(0,1,0)$ lies on the plane. Thus,

Substitute into the parallel condition:

So the plane is

Dividing by $4$,

Take point $(-2,3,-4)$ on the line. Distance from this point to the plane is

Thus the required distance is $\sqrt{14}$. This numerical result matches option A in the listed options, although the solution labels the correct option as D.