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JEE Mathematics 2025 Question with Solution

Let the point AA divide the line segment joining the points P(−1,−1,2)P(-1, -1, 2) and Q(5,5,10)Q(5, 5, 10) internally in the ratio r:1r : 1 (r>0r > 0). If OO is the origin and

(∣OQ→⋅OA→∣5)−15∣OP→×OA→∣2=10,\left( \frac{|\overrightarrow{OQ} \cdot \overrightarrow{OA}|}{5} \right) - \frac{1}{5} |\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10,

then the value of rr is:

  • A

    1414

  • B

    33

  • C

    7\sqrt{7}

  • D

    77

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Point AA divides the line segment joining P(−1,−1,2)P(-1,-1,2) and Q(5,5,10)Q(5,5,10) internally in the ratio r:1r:1. Also,

(∣OQ→⋅OA→∣5)−15∣OP→×OA→∣2=10\left( \frac{|\overrightarrow{OQ} \cdot \overrightarrow{OA}|}{5} \right) - \frac{1}{5}|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10

Find: The value of rr.

Using the section formula,

A(5r−1r+1,5r−1r+1,10r+2r+1)A\left(\frac{5r-1}{r+1},\frac{5r-1}{r+1},\frac{10r+2}{r+1}\right)

Hence,

OA→=(5r−1r+1,5r−1r+1,10r+2r+1),OP→=(−1,−1,2),OQ→=(5,5,10)\overrightarrow{OA}=\left(\frac{5r-1}{r+1},\frac{5r-1}{r+1},\frac{10r+2}{r+1}\right),\quad \overrightarrow{OP}=(-1,-1,2),\quad \overrightarrow{OQ}=(5,5,10)

Now compute the dot product:

OQ→⋅OA→=5⋅5r−1r+1+5⋅5r−1r+1+10⋅10r+2r+1\overrightarrow{OQ}\cdot \overrightarrow{OA}=5\cdot \frac{5r-1}{r+1}+5\cdot \frac{5r-1}{r+1}+10\cdot \frac{10r+2}{r+1} =25r−5+25r−5+100r+20r+1=150r+10r+1=\frac{25r-5+25r-5+100r+20}{r+1}=\frac{150r+10}{r+1}

So,

∣OQ→⋅OA→∣=∣150r+10r+1∣|\overrightarrow{OQ}\cdot \overrightarrow{OA}|=\left|\frac{150r+10}{r+1}\right|

From the solution working, the magnitude of the cross product is obtained as

∣OP→×OA→∣=6r+6r+1|\overrightarrow{OP}\times \overrightarrow{OA}|=\frac{6r+6}{r+1}

Therefore,

∣OP→×OA→∣2=(6r+6)2(r+1)2|\overrightarrow{OP}\times \overrightarrow{OA}|^2=\frac{(6r+6)^2}{(r+1)^2}

Substitute into the given equation:

15∣150r+10r+1∣−15((6r+6)2(r+1)2)=10\frac{1}{5}\left|\frac{150r+10}{r+1}\right|-\frac{1}{5}\left(\frac{(6r+6)^2}{(r+1)^2}\right)=10

Simplifying and solving, we obtain

r=7r=7

Therefore, the value of rr is 77. Hence, the correct option is D.

Section Formula with Vector Quantities

Given: OP→=⟨−1,−1,2⟩\overrightarrow{OP}=\langle -1,-1,2\rangle, OQ→=⟨5,5,10⟩\overrightarrow{OQ}=\langle 5,5,10\rangle, and AA divides PQPQ in the ratio r:1r:1. Find: The value of rr satisfying the given vector equation.

The coordinates of AA are

A(r⋅5+1⋅(−1)r+1,r⋅5+1⋅(−1)r+1,r⋅10+1⋅2r+1)A\left(\frac{r\cdot 5+1\cdot(-1)}{r+1},\frac{r\cdot 5+1\cdot(-1)}{r+1},\frac{r\cdot 10+1\cdot 2}{r+1}\right)

Thus,

A(5r−1r+1,5r−1r+1,10r+2r+1)A\left(\frac{5r-1}{r+1},\frac{5r-1}{r+1},\frac{10r+2}{r+1}\right)

Now,

OQ→⋅OA→=5⋅5r−1r+1+5⋅5r−1r+1+10⋅10r+2r+1=150r+10r+1\overrightarrow{OQ}\cdot \overrightarrow{OA}=5\cdot \frac{5r-1}{r+1}+5\cdot \frac{5r-1}{r+1}+10\cdot \frac{10r+2}{r+1}=\frac{150r+10}{r+1}

For the cross product, the solution sets up

OP→×OA→=∣i^j^k^−1−125r−1r+15r−1r+110r+2r+1∣\overrightarrow{OP}\times \overrightarrow{OA}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 2 \\ \frac{5r-1}{r+1} & \frac{5r-1}{r+1} & \frac{10r+2}{r+1} \end{vmatrix}

and then uses its magnitude in the equation.

Substituting in

(∣OQ→⋅OA→∣5)−15∣OP→×OA→∣2=10\left( \frac{|\overrightarrow{OQ} \cdot \overrightarrow{OA}|}{5} \right) - \frac{1}{5} |\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10

and solving gives

r=7r=7

Thus, the correct option is D.

Common mistakes

  • Using the section formula in the wrong order. For internal division in the ratio r:1r:1, the coordinates of AA must be formed with the opposite endpoint weights. Reversing the weights changes OA→\overrightarrow{OA} and leads to a wrong value of rr.

  • Computing ∣OP→×OA→∣2|\overrightarrow{OP} \times \overrightarrow{OA}|^2 as the cross product itself. The given expression contains the square of the magnitude, so after finding the cross product, its magnitude must be taken and then squared.

  • Dropping the modulus in ∣OQ→⋅OA→∣|\overrightarrow{OQ} \cdot \overrightarrow{OA}|. The dot product appears inside absolute value, so its sign cannot be assumed without checking.

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