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Vector Algebra question

2025 · 23 Jan · Shift 2 · Q29
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  5. /2025 · 23 Jan · Shift 2 · Q29

Vector Algebra question

2025 · 23 Jan · Shift 2 · Q29

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let the point A divide the line segment joining the points P(−1,−1,2)\mathrm{P}(-1,-1,2)P(−1,−1,2) and Q(5,5,10)\mathrm{Q}(5,5,10)Q(5,5,10) internally in the ratio r:1(r>0)r: 1(r\gt 0)r:1(r>0). If O is the origin and (OQ→⋅OA→)−15∣OP→×OA→∣2=10(\overrightarrow{\mathrm{OQ}} \cdot \overrightarrow{\mathrm{OA}})-\frac{1}{5}|\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{OA}}|^2=10(OQ​⋅OA)−51​∣OP×OA∣2=10, then the value of r is :
  1. A
    7\sqrt77​
  2. B
    14
  3. C
    7
  4. D
    3
View written solutionFree

Correct answer: C

  1. Coordinates of the given points

We have P(−1,−1,2),Q(5,5,10),O=(0,0,0).P(-1,-1,2), \qquad Q(5,5,10), \qquad O=(0,0,0).P(−1,−1,2),Q(5,5,10),O=(0,0,0).

Since AAA divides PQPQPQ internally in the ratio r:1r:1r:1, using the section formula: A=(r⋅5+(−1)r+1,r⋅5+(−1)r+1,r⋅10+2r+1).A=\left(\frac{r\cdot 5+(-1)}{r+1},\frac{r\cdot 5+(-1)}{r+1},\frac{r\cdot 10+2}{r+1}\right).A=(r+1r⋅5+(−1)​,r+1r⋅5+(−1)​,r+1r⋅10+2​). So, 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).A=(r+15r−1​,r+15r−1​,r+110r+2​).

Thus, OA→=(5r−1r+1,5r−1r+1,10r+2r+1),\overrightarrow{OA}=\left(\frac{5r-1}{r+1},\frac{5r-1}{r+1},\frac{10r+2}{r+1}\right),OA=(r+15r−1​,r+15r−1​,r+110r+2​), OQ→=(5,5,10),OP→=(−1,−1,2).\overrightarrow{OQ}=(5,5,10), \qquad \overrightarrow{OP}=(-1,-1,2).OQ​=(5,5,10),OP=(−1,−1,2).

  1. Compute OQ→⋅OA→\overrightarrow{OQ}\cdot \overrightarrow{OA}OQ​⋅OA
=5\cdot \frac{5r-1}{r+1}+5\cdot \frac{5r-1}{r+1}+10\cdot \frac{10r+2}{r+1}.$$ $$=\frac{10(5r-1)+10(10r+2)}{r+1} =\frac{50r-10+100r+20}{r+1} =\frac{150r+10}{r+1}.$$ --- 3. **Compute $\overrightarrow{OP}\times \overrightarrow{OA}$** Let $$\overrightarrow{OP}=(-1,-1,2), \qquad \overrightarrow{OA}=(a,a,b),$$ where $$a=\frac{5r-1}{r+1}, \qquad b=\frac{10r+2}{r+1}.$$ Now, $$\overrightarrow{OP}\times \overrightarrow{OA} =\begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & -1 & 2 \\ a & a & b \end{vmatrix}.$$ Expanding: $$\overrightarrow{OP}\times \overrightarrow{OA} =\hat i((-1)b-2a)-\hat j((-1)b-2a)+\hat k((-1)a-(-1)a).$$ $$=(-(b+2a),\; b+2a,\;0).$$ Hence, $$|\overrightarrow{OP}\times \overrightarrow{OA}|^2=(b+2a)^2+(b+2a)^2=2(b+2a)^2.$$ Now compute $b+2a$: $$b+2a=\frac{10r+2}{r+1}+2\cdot \frac{5r-1}{r+1} =\frac{10r+2+10r-2}{r+1} =\frac{20r}{r+1}.$$ So, $$|\overrightarrow{OP}\times \overrightarrow{OA}|^2=2\left(\frac{20r}{r+1}\right)^2 =2\cdot \frac{400r^2}{(r+1)^2} =\frac{800r^2}{(r+1)^2}.$$ Therefore, $$\frac{1}{5}|\overrightarrow{OP}\times \overrightarrow{OA}|^2 =\frac{160r^2}{(r+1)^2}.$$ --- 4. **Use the given equation** Given: $$(\overrightarrow{OQ}\cdot \overrightarrow{OA})-\frac{1}{5}|\overrightarrow{OP}\times \overrightarrow{OA}|^2=10.$$ Substitute the values: $$\frac{150r+10}{r+1}-\frac{160r^2}{(r+1)^2}=10.$$ Multiply by $(r+1)^2$: $$(150r+10)(r+1)-160r^2=10(r+1)^2.$$ Expand: $$150r^2+160r+10-160r^2=10(r^2+2r+1).$$ $$-10r^2+160r+10=10r^2+20r+10.$$ Bring all terms to one side: $$-20r^2+140r=0.$$ $$-20r(r-7)=0.$$ So, $$r=0 \quad \text{or} \quad r=7.$$ Given $r>0$, we get $$\boxed{r=7}.$$ --- 5. **Check options** - A: $\sqrt7$ ❌ - B: $14$ ❌ - C: $7$ ✅ - D: $3$ ❌ So the correct option is **C**.
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