Question ID: #312
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 $(\overrightarrow{OQ}\cdot\overrightarrow{OA}) – \frac{1}{5}|\overrightarrow{OP}\times\overrightarrow{OA}|^{2}=10$, then the value of $r$ is:
- (1) 14
- (2) 3
- (3) $\sqrt{7}$
- (4) 7
Solution:
Let the coordinates of $A$ dividing $PQ$ in ratio $r:1$ be:
$$\vec{OA} = \frac{r\vec{Q} + 1\vec{P}}{r+1}$$
$$P = (-1, -1, 2) \Rightarrow \vec{P} = -\hat{i} – \hat{j} + 2\hat{k}$$
$$Q = (5, 5, 10) \Rightarrow \vec{Q} = 5\hat{i} + 5\hat{j} + 10\hat{k}$$
$$A = \left( \frac{5r-1}{r+1}, \frac{5r-1}{r+1}, \frac{10r+2}{r+1} \right)$$
Calculate $\overline{OQ} \cdot \overline{OA}$:
$$\vec{OQ} \cdot \vec{OA} = 5\left(\frac{5r-1}{r+1}\right) + 5\left(\frac{5r-1}{r+1}\right) + 10\left(\frac{10r+2}{r+1}\right)$$
$$= \frac{1}{r+1} [25r – 5 + 25r – 5 + 100r + 20]$$
$$= \frac{150r + 10}{r+1} = \frac{10(15r+1)}{r+1}$$
Calculate $\overline{OP} \times \overline{OA}$:
Since $A$ lies on the line passing through $P$, the vectors $\vec{OP}$ and $\vec{OA}$ are not necessarily collinear, but $A, P, Q$ are collinear.
Let’s compute the cross product directly or simplify.
$\vec{OA} = \frac{r\vec{Q} + \vec{P}}{r+1}$.
$$\vec{OP} \times \vec{OA} = \vec{OP} \times \left(\frac{r\vec{Q} + \vec{P}}{r+1}\right)$$
$$= \frac{r}{r+1}(\vec{OP} \times \vec{Q}) + \frac{1}{r+1}(\vec{OP} \times \vec{P})$$
Since $\vec{OP} \times \vec{P} = 0$:
$$\vec{OP} \times \vec{OA} = \frac{r}{r+1}(\vec{OP} \times \vec{Q})$$
Now compute $\vec{OP} \times \vec{Q}$:
$$\vec{OP} \times \vec{Q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 2 \\ 5 & 5 & 10 \end{vmatrix}$$
$$= \hat{i}(-10 – 10) – \hat{j}(-10 – 10) + \hat{k}(-5 – (-5))$$
$$= -20\hat{i} + 20\hat{j} + 0\hat{k}$$
Magnitude square $|\vec{OP} \times \vec{Q}|^2 = (-20)^2 + (20)^2 = 400 + 400 = 800$.
So,
$$|\vec{OP} \times \vec{OA}|^2 = \frac{r^2}{(r+1)^2} (800)$$
Substitute these into the given equation:
$$(\overline{OQ}\cdot\overline{OA}) – \frac{1}{5}|\overline{OP}\times\overline{OA}|^{2}=10$$
$$\frac{10(15r+1)}{r+1} – \frac{1}{5} \left[ \frac{800r^2}{(r+1)^2} \right] = 10$$
Divide by 10:
$$\frac{15r+1}{r+1} – \frac{16r^2}{(r+1)^2} = 1$$
Multiply by $(r+1)^2$:
$$(15r+1)(r+1) – 16r^2 = (r+1)^2$$
$$15r^2 + 15r + r + 1 – 16r^2 = r^2 + 2r + 1$$
$$-r^2 + 16r + 1 = r^2 + 2r + 1$$
$$2r^2 – 14r = 0$$
$$2r(r – 7) = 0$$
Since $r > 0$, we have $r = 7$.
Ans. (4)
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