Let the point A divide the line segment joining the points $\mathrm{P}(-1,-1,2)$ and $\mathrm{Q}(5,5,10)$ internally in the ratio $r: 1(r>0)$. If $O$ is the origin and $(\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{1}{5}|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^{2}=10$, then the value of r is :

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Vector Algebra

Let the point A divide the line segment joining the points

P (- 1, - 1, 2)

and

Q (5, 5, 10)

internally in the ratio

r : 1 (r > 0)

. If

O

is the origin and

(OQ \cdot OA) - \frac{1}{5} ∣ OP \times OA ∣^{2} = 10

, then the value of r is :

Answer: (D)

Since A divides PQ in ratio

r : 1

with

P (- 1, - 1, 2)

Q (5, 5, 10)

O A = \frac{O P + r O Q}{r + 1} = (\frac{5 r - 1}{r + 1}, \frac{5 r - 1}{r + 1}, \frac{10 r + 2}{r + 1})

O Q \cdot O A = \frac{5 ( 5 r - 1 ) + 5 ( 5 r - 1 ) + 10 ( 10 r + 2 )}{r + 1} = \frac{150 r + 10}{r + 1}

For

O P \times O A

, with

O P = (- 1, - 1, 2)

O P \times O A = \frac{1}{r + 1} (- 20 r, 20 r, 0)

∣ O P \times O A ∣^{2} = \frac{800 r ^{2}}{( r + 1 ) ^{2}}

Substituting into

O Q \cdot O A - \frac{1}{5} ∣ O P \times O A ∣^{2} = 10

\frac{150 r + 10}{r + 1} - \frac{160 r ^{2}}{( r + 1 ) ^{2}} = 10

Multiplying through by

(r + 1)^{2}

(150 r + 10) (r + 1) - 160 r^{2} = 10 (r + 1)^{2}

150 r^{2} + 160 r + 10 - 160 r^{2} = 10 r^{2} + 20 r + 10

- 10 r^{2} + 160 r = 10 r^{2} + 20 r

20 r^{2} - 140 r = 0 \Rightarrow 20 r (r - 7) = 0

Since

r > 0

r = 7

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option