Question

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 $(\overline{OQ} \cdot \overline{OA}) - \frac{1}{5}|\overline{OP} \times \overline{OA}|^2 = 10$ then the value of r

Answer 1

7

Answer 2

The position vector of point A is given by $\vec{OA} = \frac{r\vec{OQ} + \vec{OP}}{r+1}$.

We calculate the dot product: $\vec{OQ} \cdot \vec{OA} = \langle 5, 5, 10 \rangle \cdot \frac{\langle 5r-1, 5r-1, 10r+2 \rangle}{r+1} = \frac{150r+10}{r+1}$.

We calculate the cross product: $\vec{OP} \times \vec{OA} = \frac{r}{r+1}(\vec{OP} \times \vec{OQ}) = \frac{r}{r+1} \langle -20, 20, 0 \rangle$.

The squared magnitude of the cross product is: $|\vec{OP} \times \vec{OA}|^2 = \left(\frac{r}{r+1}\right)^2 ((-20)^2 + 20^2) = \frac{800r^2}{(r+1)^2}$.

Substitute into the given equation: $\frac{150r+10}{r+1} - \frac{1}{5} \left(\frac{800r^2}{(r+1)^2}\right) = 10$ $\frac{150r+10}{r+1} - \frac{160r^2}{(r+1)^2} = 10$

Multiplying by $(r+1)^2$: $(150r+10)(r+1) - 160r^2 = 10(r+1)^2$ $150r^2 + 160r + 10 - 160r^2 = 10(r^2 + 2r + 1)$ $-10r^2 + 160r + 10 = 10r^2 + 20r + 10$ $20r^2 - 140r = 0$ $20r(r-7) = 0$

Since $r>0$, we have $r=7$.