JEE Mathematics 2023 Question with Solution

Given: Point $P(1, -2, 3)$, and the line passing through $A(4, 5, 8)$ and $B(1, -7, 5)$ as used in the solution.

Find: The distance of the foot of perpendicular $N$ from the plane.

First find the direction vector of the line:

$$\vec{AB} = B - A = (1 - 4, -7 - 5, 5 - 8) = (-3, -12, -3)$$

So the parametric equations are:

$$x = 4 - 3t, \quad y = 5 - 12t, \quad z = 8 - 3t$$

Now,

$$\vec{AP} = P - A = (1 - 4, -2 - 5, 3 - 8) = (-3, -7, -5)$$

For the foot of the perpendicular, use projection:

$$t = \frac{\vec{AP} \cdot \vec{AB}}{\vec{AB} \cdot \vec{AB}} = \frac{(-3)(-3) + (-7)(-12) + (-5)(-3)}{(-3)^2 + (-12)^2 + (-3)^2} = \frac{108}{162} = \frac{2}{3}$$

Substitute $t = \frac{2}{3}$ in the line:

$$x = 4 - 3\left(\frac{2}{3}\right) = 2, \quad y = 5 - 12\left(\frac{2}{3}\right) = -3, \quad z = 8 - 3\left(\frac{2}{3}\right) = 6$$

Hence,

$$N(2, -3, 6)$$

Now use the plane form written in the solution:

$$2x - 2y + z + 5 = 0$$

Distance from $N(2, -3, 6)$ to the plane is:

$$D = \frac{|2(2) - 2(-3) + 1(6) + 5|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|4 + 6 + 6 + 5|}{\sqrt{9}} = \frac{21}{3} = 7$$

Therefore, the distance of $N$ from the plane is $7$, so the correct option is B.

Note: The solution uses $B(1, -7, 5)$ and plane $2x - 2y + z + 5 = 0$, while the question text shows $B(1, -7, -5)$ and $2x - 2y + z = 5$. The extracted answer follows the solution, which is the primary source.

Given: A point $P$ and a line through two points, then a plane.

Find: First the foot of perpendicular $N$ on the line, then its distance from the plane.

The method has two stages:

1. Find $N$ using vector projection.
2. Use the point-to-plane distance formula.

For the line through $A(4, 5, 8)$ and $B(1, -7, 5)$,

$$\vec{AB} = (-3, -12, -3)$$

From $A$ to $P(1, -2, 3)$,

$$\vec{AP} = (-3, -7, -5)$$

The projection parameter is:

$$t = \frac{\vec{AP} \cdot \vec{AB}}{|\vec{AB}|^2}$$

Compute the numerator:

$$\vec{AP} \cdot \vec{AB} = 9 + 84 + 15 = 108$$

Compute the denominator:

$$|\vec{AB}|^2 = 9 + 144 + 9 = 162$$

Hence,

$$t = \frac{108}{162} = \frac{2}{3}$$

So the foot point is:

$$N = A + t\vec{AB} = (4, 5, 8) + \frac{2}{3}(-3, -12, -3) = (2, -3, 6)$$

Now for the plane $2x - 2y + z + 5 = 0$, the distance formula gives:

$$D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Substitute $A = 2$, $B = -2$, $C = 1$, $D = 5$ and $N(2, -3, 6)$:

$$D = \frac{|2\cdot 2 + (-2)(-3) + 1\cdot 6 + 5|}{\sqrt{4 + 4 + 1}} = \frac{21}{3} = 7$$

Therefore, the required distance is $7$ and the correct option is B.