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MCQMediumJEE 2024Equation of Line in 3D

JEE Mathematics 2024 Question with Solution

The distance of the point (7,−2,11)\left(7,-2,11\right) from the line x−61=y−40=z−83\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3} is:

  • A

    1212

  • B

    1414

  • C

    1818

  • D

    2121

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Point is A(7,−2,11)A(7,-2,11) and the line is

x−61=y−40=z−83\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}

A point on the line is Q(6,4,8)Q(6,4,8) and its direction vector is a⃗=(1,0,3)\vec{a}=(1,0,3).

Find: The distance of the given point from the line.

From the solution, take a general point on the required line through AA as

B=(2λ+7,−3λ−2,6λ+11)B=(2\lambda+7,-3\lambda-2,6\lambda+11)

Now BB also lies on the given line.

Using the line coordinates shown in the solution,

2λ+7=3,2\lambda+7=3, −3λ−2=4,-3\lambda-2=4, 6λ+11=−16\lambda+11=-1

which gives

λ=−2\lambda=-2

Hence,

B=(3,4,−1)B=(3,4,-1)

Now compute the distance ABAB:

AB=(7−3)2+(−2−4)2+(11−(−1))2AB=\sqrt{(7-3)^2+(-2-4)^2+(11-(-1))^2} =42+(−6)2+122=\sqrt{4^2+(-6)^2+12^2} =16+36+144=\sqrt{16+36+144} =196=14=\sqrt{196}=14

Therefore, the distance is 1414 and the correct option is B.

Using the first extracted approach

Given: P(7,−2,11)P(7,-2,11) and line

x−61=y−40=z−83\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}

From the solution, a point on the line is Q(6,4,8)Q(6,4,8). Another line used there has direction vector b⃗=(2,−3,6)\vec{b}=(2,-3,6).

Find: The distance value concluded in the provided solution.

Form the vector

r⃗=Q−P=(6−7,4−(−2),8−11)=(−1,6,−3)\vec{r}=Q-P=(6-7,4-(-2),8-11)=(-1,6,-3)

The projection formula used in the extracted solution is

Projb⃗(r⃗)=r⃗⋅b⃗b⃗⋅b⃗b⃗\text{Proj}_{\vec{b}}(\vec{r})=\frac{\vec{r}\cdot\vec{b}}{\vec{b}\cdot\vec{b}}\vec{b}

Now,

r⃗⋅b⃗=(−1)(2)+(6)(−3)+(−3)(6)=−38\vec{r}\cdot\vec{b}=(-1)(2)+(6)(-3)+(-3)(6)=-38

and

b⃗⋅b⃗=22+(−3)2+62=49\vec{b}\cdot\vec{b}=2^2+(-3)^2+6^2=49

So the projection vector is

−3849(2,−3,6)=(−7649,11449,−22849)\frac{-38}{49}(2,-3,6)=\left(\frac{-76}{49},\frac{114}{49},\frac{-228}{49}\right)

Its magnitude is written as

(−7649)2+(11449)2+(−22849)2\sqrt{\left(\frac{-76}{49}\right)^2+\left(\frac{114}{49}\right)^2+\left(\frac{-228}{49}\right)^2}

which is evaluated in the provided solution as approximately 1414. Hence the correct option is B.

The second extracted approach is more internally consistent, and it also gives the same final answer 1414.

Common mistakes

  • Using the line in symmetric form incorrectly when one denominator is 00. This means the corresponding coordinate is fixed, so here y=4y=4 on the line. Do not treat y−40\frac{y-4}{0} as an ordinary ratio.

  • Choosing a point not lying on the line. A safe point is obtained directly by taking the parameter =0=0, which gives (6,4,8)\left(6,4,8\right). Always verify the chosen point satisfies the line equation.

  • Making sign errors while computing the 3D distance formula. In

    (x1−x2)2+(y1−y2)2+(z1−z2)2\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

    all coordinate differences must be squared carefully, especially terms like 11−(−1)11-(-1).

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