Let a line L pass through the point P(2, 3, 1) and be parallel to… — Solution | JEE Mathematics

NVAMediumJEE 2023Equation of Line in 3D

Let a line L pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0$ i.e. $x - y + 2z = 0$ . If the distance of L from the point $(5, 3, 8)$ is $\alpha$ , then $3\alpha^2$ is equal to ........

Answer

Correct answer:158

Step-by-step solution

Standard Method

Given: The line L passes through $P(2,3,1)$ and is parallel to the intersection of the planes $x+3y-2z-2=0$ and $x-y+2z=0$ .

Find: The value of $3\alpha^2$ , where $\alpha$ is the distance of the point $(5,3,8)$ from L.

The direction ratios of the line parallel to the intersection of the two planes are obtained from the cross product of their normals:

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ 1 & -1 & 2 \end{vmatrix} = 4\hat{i} - 4\hat{j} - 4\hat{k}

So a direction vector of L is proportional to

(1,-1,-1)

Hence the equation of the line is

\frac{x-2}{1} = \frac{y-3}{-1} = \frac{z-1}{-1}

Let $Q=(5,3,8)$ and let the foot of the perpendicular from $Q$ to the line be $R$ . A general point on the line is

R \equiv (k+2,-k+3,-k+1)

Then the direction ratios of $QR$ are

(k-3,-k,-k-7)

Since $QR \perp L$ , their dot product with direction vector $(1,-1,-1)$ is zero:

(1)(k-3)+(-1)(-k)+(-1)(-k-7)=0

3k+4=0

k=-\frac{4}{3}

Substituting this value of $k$ , we get

QR = \left(-\frac{13}{3},\frac{4}{3},-\frac{17}{3}\right)

Therefore,

\alpha^2 = \left(\frac{13}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(\frac{17}{3}\right)^2

\alpha^2 = \frac{169+16+289}{9} = \frac{474}{9} = \frac{158}{3}

Hence,

3\alpha^2 = 158

Therefore, the required answer is $158$ .

Using foot of perpendicular parametrically

Given: A line through $P(2,3,1)$ parallel to the line of intersection of two planes.

Find: $3\alpha^2$ .

The normals of the planes are $\vec{n}_1=(1,3,-2)$ and $\vec{n}_2=(1,-1,2)$ . Their cross product gives the direction of the required line:

\vec{d}=\vec{n}_1 \times \vec{n}_2 = (4,-4,-4)

So we may use

\vec{d}=(1,-1,-1)

Distance from a point to a line in vector form

Given: Point on line $P=(2,3,1)$ , external point $Q=(5,3,8)$ , and direction vector $\vec{d}=(1,-1,-1)$ .

Find: $3\alpha^2$ .

Using the vector formula for distance from a point to a line,

\alpha^2 = |Q-P|^2 - \frac{\big((Q-P)\cdot d\big)^2}{|d|^2}

Here,

Q-P = (3,0,7)

So,

|Q-P|^2 = 3^2+0^2+7^2 = 58

(Q-P)\cdot d = 3+0-7 = -4

|d|^2 = 1+1+1 = 3

Thus,

\alpha^2 = 58 - \frac{16}{3} = \frac{158}{3}

Hence,

3\alpha^2 = 158

So the correct answer is $158$ .

Common mistakes

Using the wrong direction vector for the intersection line. The line is parallel to the intersection of two planes, so its direction must be the cross product of the plane normals, not one of the normals themselves.
Writing the symmetric equation with an incorrect third denominator. A sign error in the direction ratios changes the entire line and gives a wrong perpendicular distance.
Forgetting the perpendicularity condition $QR \perp L$ . The foot of the perpendicular is not any point on the line; the dot product with the line direction must be zero.

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