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MCQMediumJEE 2023Equation of Plane

JEE Mathematics 2023 Question with Solution

Let P be the plane, passing through the point (1,−1,−5)(1, -1, -5) and perpendicular to the line joining the points (4,1,−3)(4, 1, -3) and (2,4,3)(2, 4, 3). Then the distance of P from the point (3,−2,2)(3, -2, 2) is:

  • A

    66

  • B

    44

  • C

    55

  • D

    77

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The plane passes through (1,−1,−5)(1, -1, -5) and is perpendicular to the line joining (4,1,−3)(4, 1, -3) and (2,4,3)(2, 4, 3).

Find: The distance of this plane from the point (3,−2,2)(3, -2, 2).

The direction ratios of the line joining (4,1,−3)(4, 1, -3) and (2,4,3)(2, 4, 3) are

(2−4,  4−1,  3−(−3))=(−2,  3,  6)(2-4,\;4-1,\;3-(-3)) = (-2,\;3,\;6)

So a normal vector to the plane can be taken as (2,−3,−6)(2, -3, -6).

Hence the equation of the plane is

2(x−1)−3(y+1)−6(z+5)=02(x-1) - 3(y+1) - 6(z+5) = 0

Or,

2x−3y−6z=352x - 3y - 6z = 35

Now the distance of the point (3,−2,2)(3, -2, 2) from the plane 2x−3y−6z−35=02x - 3y - 6z - 35 = 0 is

∣2(3)−3(−2)−6(2)−35∣22+(−3)2+(−6)2\frac{|2(3) - 3(-2) - 6(2) - 35|}{\sqrt{2^2 + (-3)^2 + (-6)^2}} =∣6+6−12−35∣4+9+36= \frac{|6 + 6 - 12 - 35|}{\sqrt{4+9+36}} =357=5= \frac{35}{7} = 5

Therefore, the distance is 55 and the correct option is C.

Distance Formula Expansion

Given: Plane through (1,−1,−5)(1, -1, -5) and perpendicular to the line through (4,1,−3)(4, 1, -3) and (2,4,3)(2, 4, 3).

Find: Distance from (3,−2,2)(3, -2, 2) to the plane.

Because the plane is perpendicular to the given line, the line's direction vector becomes the normal vector of the plane. Using the two points on the line,

n⃗=(2−4,  4−1,  3+3)=(−2,  3,  6)\vec{n} = (2-4,\;4-1,\;3+3) = (-2,\;3,\;6)

An equivalent normal vector is

(2,−3,−6)(2, -3, -6)

Using point-normal form with point (1,−1,−5)(1, -1, -5),

2(x−1)−3(y+1)−6(z+5)=02(x-1) - 3(y+1) - 6(z+5) = 0

Expanding,

2x−2−3y−3−6z−30=02x - 2 - 3y - 3 - 6z - 30 = 0 2x−3y−6z−35=02x - 3y - 6z - 35 = 0

Substitute (x1,y1,z1)=(3,−2,2)(x_1,y_1,z_1) = (3,-2,2) into the point-to-plane distance formula

Distance=∣Ax1+By1+Cz1+D∣A2+B2+C2\text{Distance} = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2+B^2+C^2}}

where A=2,  B=−3,  C=−6,  D=−35A=2,\;B=-3,\;C=-6,\;D=-35.

So,

Distance=∣2⋅3+(−3)(−2)+(−6)(2)−35∣22+(−3)2+(−6)2\text{Distance} = \frac{|2\cdot 3 + (-3)(-2) + (-6)(2) - 35|}{\sqrt{2^2+(-3)^2+(-6)^2}} =∣6+6−12−35∣49= \frac{|6+6-12-35|}{\sqrt{49}} =357=5= \frac{35}{7} = 5

Therefore, the required distance is 55.

Common mistakes

  • Using the line's direction vector incorrectly. Since the plane is perpendicular to the line, that direction vector becomes the normal to the plane. Do not treat it as a vector lying in the plane.

  • Making an error while finding direction ratios from the two given points, especially the third coordinate: 3−(−3)=63-(-3)=6, not 00 or −6-6. Compute each component carefully.

  • Using the point-to-plane distance formula with the wrong constant term. After expanding the plane equation, write it in the form Ax+By+Cz+D=0Ax+By+Cz+D=0 before substitution.

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