If the distance of the point P(4,,), <0, from the line r = 4 i- k+ (2 i+3 k), R along a line with direction ratios 3,-1,0 is 13 10, then ^2+ ^2 is equal to.

Given: A point on the given line is A(4,0,-1). The direction vector of the line is d1=(2,0,3). The direction along which distance is measured is d2=(3,-1,0) with magnitude | d2|= 10. The point is P(4,,). Find: The value of ^2+ ^2. From the solution working, the vector joining the point on the line to P is AP=(4 -4,, +1) Distance along direction d2 is obtained by projection: | AP d2|| d2|= 13 10 So, | AP d2|=13 Now, |(4 -4)3+ (-1)|=13 |11 -12|=13 Using the branch taken in the extracted solution, 11 -12=-13 =- 111 Also, the perpendicularity condition used in the solution is AP d1=0 Therefore, (4 -4)2+( +1)3=0 Substituting =- 111, =- 1711 Hence, ^2+ ^2= 1121+ 289121 = 290121 Therefore, the required value is 290121. The solution concludes this value, which disagrees with the answer key 2.4.

If the distance of the point P(4,,), <0, from the line r = 4 i- k… — Solution | JEE Mathematics

NVAMediumJEE 2026Equation of Line in 3D

If the distance of the point $P(4\alpha,\alpha,\beta)$ , $\beta<0$ , from the line

\vec r = 4\hat i-\hat k+\mu(2\hat i+3\hat k),\ \mu\in\mathbb{R}

along a line with direction ratios $3,-1,0$ is $\dfrac{13}{\sqrt{10}}$ , then $\alpha^2+\beta^2$ is equal to _____.

Answer

Correct answer:2.4

Step-by-step solution

Standard Method

Given: A point on the given line is $A(4,0,-1)$ . The direction vector of the line is $\vec d_1=(2,0,3)$ . The direction along which distance is measured is $\vec d_2=(3,-1,0)$ with magnitude $|\vec d_2|=\sqrt{10}$ . The point is $P(4\alpha,\alpha,\beta)$ .

Find: The value of $\alpha^2+\beta^2$ .

From the solution working, the vector joining the point on the line to $P$ is

\vec{AP}=(4\alpha-4,\alpha,\beta+1)

Distance along direction $\vec d_2$ is obtained by projection:

\frac{|\vec{AP}\cdot \vec d_2|}{|\vec d_2|}=\frac{13}{\sqrt{10}}

So,

|\vec{AP}\cdot \vec d_2|=13

Now,

|(4\alpha-4)3+\alpha(-1)|=13

|11\alpha-12|=13

Using the branch taken in the extracted solution,

11\alpha-12=-13

\alpha=-\frac{1}{11}

Also, the perpendicularity condition used in the solution is

\vec{AP}\cdot\vec d_1=0

Therefore,

(4\alpha-4)2+(\beta+1)3=0

Substituting $\alpha=-\frac{1}{11}$ ,

\beta=-\frac{17}{11}

Hence,

\alpha^2+\beta^2=\frac{1}{121}+\frac{289}{121}

=\frac{290}{121}

Therefore, the required value is $\frac{290}{121}$ . The solution concludes this value, which disagrees with the answer key $2.4$ .

Common mistakes

Using the shortest distance from a point to a line instead of the distance measured along the given direction. This is wrong because the question asks for projection along direction ratios $3,-1,0$ . Instead, project the joining vector on $\vec d_2=(3,-1,0)$ .
Forgetting the condition $\vec{AP}\cdot\vec d_1=0$ for the relevant joining vector. This is wrong because the point on the line connected to $P$ must be chosen consistently with the line's direction. Use the perpendicularity relation exactly as shown in the solution working.
Dropping the modulus in $|11\alpha-12|=13$ too early. This is wrong because both algebraic branches are possible before applying the given condition. First solve the absolute value equation, then use the stated condition to select the branch used in the solution.

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